Prove $\phi(p^n) = (p-1)p^{n-1}$ for $p$ prime

Question

Is there a proper proof of the following property:

Let $p$ be a prime number. The number of invertible elements in $\mathbb{Z}/p^n\mathbb{Z}$ is $(p-1)p^{n-1}$.

Answer 1

An element $\overline{a}$ (where $0\leq a \leq n-1)$ of the ring $\mathbb{Z}/n$ is invertible precisely when $a$ is coprime to $n$ by a standard result of elementary number theory. The number of positive integers less than or equal to $n$ which are coprime to $n$ is given by the euler-phi function.

Hence it suffices to compute $\varphi(p^n)$. You can show that $\varphi(p^n)=(p-1)p^{n-1}$ by noting that the only postive integers less than or equal to $p^n$ which are not coprime to it are the multiples of $p$ namely $kp$ for $k=1,\dotsc, p^{n-1}$. Hence $$\varphi(p^n)=p^n-p^{n-1}=(p-1)p^{n-1}.$$

Answer 2

Let $G=\mathbb{Z}/p^{n}\mathbb{Z}$.

Lemma. For $a,n\in\mathbb{N}$, $ax\pmod{m}=1$ has a solution if and only if $\gcd(a,m)=1$.

Proof. $ab\pmod{m}=1\iff m\mid ab-1\iff\exists k\in\mathbb{N}$ such that $mk=ab-1\iff1=a(b)+m(-k)\iff\gcd(a,m)=1$.//

Now, by our lemma we know that the invertible elements in $G$ are precisely those that are relatively prime to $p^{n}$. If you are familiar with the Euler-totient function, then you know that the number of such elements is $\varphi(p^{n})=p^{n}-p^{n-1}=(p-1)p^{n-1}$.