Here is a proof that I'm copypasting from my old homework. It proves a bit more (see Theorem 3) and does not use the explicit formula for $\phi$ through the prime factorization.
Let $\phi$ denote Euler's totient function.
Lemma 1. Let $m$ and $k$ be positive integers. Let $s\in\mathbb{Z}$.
(a) We have
\begin{align}
& \left( \text{the number of integers among }ms+1,\ ms+2,\ \ldots
,\ ms+m\text{ that are coprime to }m\right) \nonumber\\
& =\phi\left( m\right) .
\label{darij1.p11.claim1}
\tag{1}
\end{align}
(b) Let $x\in\left\{ 1,2,\ldots,m\right\} $. The integer $ms+x$ is
coprime to $m$ if and only if $x$ is coprime to $m$.
Proof of Lemma 1. (b) If $d$ is a common divisor of $ms+x$ and $m$, then
$d$ must also divide $x$ (since $d\mid ms+x$ and $d\mid m$, so that
\begin{equation}
x=\underbrace{\left( ms+x\right) }_{\substack{\equiv0\mod
d\\\text{(since }d\mid ms+x\text{)}}}-\underbrace{m}_{\substack{\equiv
0\mod d\\\text{(since }d\mid m\text{)}}}s\equiv
0-0s=0\mod d,
\end{equation}
so that $d\mid x$), and thus $d$ must be a common divisor of $x$ and $m$. In
other words, each common divisor of $ms+x$ and $m$ must be a common divisor of
$x$ and $m$.
If $d$ is a common divisor of $x$ and $m$, then $d$ must also divide $ms+x$
(since $d\mid x$ and $d\mid m$, so that
\begin{equation}
\underbrace{m}_{\substack{\equiv0\mod d\\\text{(since }d\mid
m\text{)}}}s+\underbrace{x}_{\substack{\equiv0\mod
d\\\text{(since }d\mid x\text{)}}}\equiv0s+0=0\mod d,
\end{equation}
so that $d\mid ms+x$), and thus $d$ must be a common divisor of $ms+x$ and
$m$. In other words, each common divisor of $x$ and $m$ must be a common
divisor of $ms+x$ and $m$.
We have thus shown that:
Combining these two statements, we conclude that the common divisors of $ms+x$
and $m$ are exactly the common divisors of $x$ and $m$. Thus, the greatest
common divisor of $ms+x$ and $m$ is the greatest common divisor of $x$ and
$m$. In other words, $\gcd\left( ms+x,m\right) =\gcd\left( x,m\right) $.
Now, we have the following equivalence of statements:
\begin{align*}
\left( ms+x\text{ is coprime to }m\right) \ & \Longleftrightarrow\ \left(
\underbrace{\gcd\left( ms+x,m\right) }_{=\gcd\left( x,m\right) }=1\right)
\\
& \Longleftrightarrow\ \left( \gcd\left( x,m\right) =1\right)
\ \Longleftrightarrow\ \left( x\text{ is coprime to }m\right) .
\end{align*}
This proves Lemma 1 (b).
(a) We have
\begin{align*}
& \left( \text{the number of integers among }ms+1,\ ms+2,\ \ldots
,\ ms+m\text{ that are coprime to }m\right) \\
& =\left\vert \left\{ x\in\left\{ 1,2,\ldots,m\right\} \ \mid
\ \underbrace{ms+x\text{ is coprime to }m}_{\substack{\text{this is equivalent
to}\\x\text{ being coprime to }m\\\text{(by Lemma 1 (b))}}}\right\}
\right\vert \\
& =\left\vert \left\{ x\in\left\{ 1,2,\ldots,m\right\} \ \mid\ x\text{ is
coprime to }m\right\} \right\vert \\
& =\left( \text{the number of integers among }1,\ 2,\ \ldots,\ m\text{ that
are coprime to }m\right) \\
& =\left( \text{the number of }x\in\left\{ 1,2,\ldots,m\right\} \text{
that are coprime to }m\right) \\
& =\phi\left( m\right)
\end{align*}
(since $\phi\left( m\right) $ is defined as the number of $x\in\left\{
1,2,\ldots,m\right\} $ that are coprime to $m$). This proves Lemma 1 (a).
$\blacksquare$
Lemma 2. Let $m$ and $k$ be positive integers. Then, the number of
positive integers $\leq mk$ that are coprime to $m$ is $k\phi\left( m\right)
$.
Proof of Lemma 2. We have
\begin{align*}
& \left( \text{the number of positive integers }\leq mk\text{ that are
coprime to }m\right) \\
& =\left( \text{the number of integers among }1,\ 2,\ \ldots,\ mk\text{ that
are coprime to }m\right) \\
& =\sum_{s=0}^{k-1}\underbrace{\left( \text{the number of integers among
}ms+1,\ ms+2,\ \ldots,\ ms+m\text{ that are coprime to }m\right)
}_{\substack{=\phi\left( m\right) \\\text{(by \eqref{darij1.p11.claim1})}}}\\
& \qquad\qquad\left(
\begin{array}
[c]{c}
\text{since each of the integers }1,\ 2,\ \ldots,\ mk\text{ belongs to exactly
one}\\
\text{of the }k\text{ lists }\left( 1,\ 2,\ \ldots,\ m\right) ,\ \left(
m+1,\ m+2,\ \ldots,\ 2m\right) ,\ \ldots,\\
\left( m\left( k-1\right) +1,\ m\left( k-1\right) +2,\ \ldots
,\ mk\right) \text{; in other words, each}\\
\text{of the integers }1,\ 2,\ \ldots,\ mk\text{ belongs to exactly one of the
}k\\
\text{lists }\left( ms+1,\ ms+2,\ \ldots,\ ms+m\right) \text{ with }
s\in\left\{ 0,1,\ldots,k-1\right\}
\end{array}
\right) \\
& =\sum_{s=0}^{k-1}\phi\left( m\right) =k\phi\left( m\right) .
\end{align*}
This proves Lemma 2. $\blacksquare$
Theorem 3. Let $m$ and $n$ be two positive integers. Assume that every
prime that divides $n$ also divides $m$. Then:
(a) For every integer $x$, we have the following logical equivalence:
\begin{equation}
\left( x\text{ is coprime to }m\right) \Longleftrightarrow\left( x\text{ is
coprime to }mn\right) .
\label{darij1.p12.c1}
\tag{2}
\end{equation}
(b) We have $\phi\left( nm\right) =n\phi\left( m\right) $.
Proof of Theorem 3. (a) Let $x$ be an integer.
Proof of the $\Longrightarrow$ direction of \eqref{darij1.p12.c1}:
Assume
that $x$ is coprime to $m$. Thus, $\gcd\left( x,m\right) =1$. Hence, no
prime can divide both $m$ and $x$. Now, we shall prove that $x$ is coprime to
$mn$. Indeed, assume the contrary. That is, $x$ is not coprime to $mn$; thus,
$\gcd\left( x,mn\right) >1$. Hence, there exists a prime $p$ which divides
$\gcd\left( x,mn\right) $. Consider this $p$. Then, $p\mid\gcd\left(
x,mn\right) \mid mn$. Since $p$ is a prime, this yields that we have $p\mid
m$ or $p\mid n$. If we had $p\mid m$, then the prime $p$ would divide both $m$
and $x$ (because $p\mid m$ and $p\mid\gcd\left( x,mn\right) \mid x$); this
would contradict the fact that no prime can divide both $m$ and $x$. Hence, we
cannot have $p\mid m$. Thus, we must have $p\mid n$ (since we have $p\mid m$
or $p\mid n$). Since $p$ is a prime, we thus conclude that $p\mid m$ (due to
our assumption that every prime that divides $n$ also divides $m$). This
contradicts the fact that we cannot have $p\mid m$. This contradiction shows
that our assumption was false. Hence, we have shown that $x$ is coprime to
$mn$. The $\Longrightarrow$ direction of \eqref{darij1.p12.c1} is thus proven.
Proof of the $\Longleftarrow$ direction of \eqref{darij1.p12.c1}:
Assume
that $x$ is coprime to $mn$. Thus, $\gcd\left( x,mn\right) =1$. Now, we
shall prove that $x$ is coprime to $m$. Indeed, assume the contrary. That is,
$x$ is not coprime to $m$; thus, $\gcd\left( x,m\right) >1$. Hence, there
exists a prime $p$ which divides $\gcd\left( x,m\right) $. Consider this
$p$. Then, $p\mid\gcd\left( x,m\right) \mid x$ and $p\mid\gcd\left(
x,m\right) \mid m\mid mn$. Hence, $p$ is a common divisor of the numbers $x$
and $mn$, and therefore divides their greatest common divisor. In other words,
$p\mid\gcd\left( x,mn\right) =1$. So we have $p\mid1$. This is absurd, since
$p$ is a prime. This contradiction shows that our assumption was false. Hence,
we have shown that $x$ is coprime to $m$. The $\Longleftarrow$ direction of
\eqref{darij1.p12.c1} is thus proven.
Now, with both directions of \eqref{darij1.p12.c1} being proven, the proof of
\eqref{darij1.p12.c1} is complete. In other words, Theorem 3 (a) is proven.
(b) Lemma 2 (applied to $k=n$) yields that the number of positive integers
$\leq mn$ that are coprime to $m$ is $n\phi\left( m\right) $. Thus,
\begin{align*}
n\phi\left( m\right) & =\left( \text{the number of positive integers
}\leq mn\text{ that are coprime to }m\right) \\
& =\left( \text{the number of }x\in\left\{ 1,2,\ldots,mn\right\} \text{
such that }\underbrace{x\text{ is coprime to }m}
_{\substack{\Longleftrightarrow\left( x\text{ is coprime to }mn\right)
\\\text{(due to \eqref{darij1.p12.c1})}}}\right) \\
& =\left( \text{the number of }x\in\left\{ 1,2,\ldots,mn\right\} \text{
such that }x\text{ is coprime to }mn\right) \\
& =\phi\left( mn\right)
\end{align*}
(since $\phi\left( mn\right) $ is defined as the number of $x\in\left\{
1,2,\ldots,mn\right\} $ such that $x$ is coprime to $mn$). This proves Theorem
3 (b). $\blacksquare$
Corollary 4. Let $m$ and $n$ be two positive integers such that $n\mid m$.
Then, $\phi\left( nm\right) =n\phi\left( m\right) $.
Proof of Corollary 4. Every prime that divides $n$ also divides $m$ (since
$n\mid m$). Hence, Theorem 3 (b) yields $\phi\left( nm\right)
=n\phi\left( m\right) $. This proves Corollary 4. $\blacksquare$
Corollary 4 is your question (with $d$ and $n$ renamed as $n$ and $m$).
