If $p$ and $q$ are distinct primes and $a$ be any integer then $a^{pq} -a^q -a^p +a$ is divisible by $pq$.
Factorising we get $a^{pq} -a^q -a^p +a =a^p(a^q -1) - a(a^{q-1}-1)$ and we know $p \mid a^{p-1}-1$ and $q \mid a^{q-1}-1$.
I can't proceed further with the proof. Please Help!
Hint Look at this $\pmod{p}$ and $\pmod{q}$. If you prove that your expression is $0$ in both modular arithmetics, you are done.
$\pmod{p}$ you have $a^p \equiv a \pmod{p}$ therefore you also have $$(a^p)^q \equiv a^q \pmod{p}$$
$\pmod{q}$ you can do a similar computation.
For prime $q,$ $$a^{pq}-a^q-a^p+a=[(\underbrace{a^p})^q-(\underbrace{a^p})]-[a^q-a]$$
Now by Fermat's Little Theorem, $b^q\equiv b\pmod q$ where $b$ is any integer
Set $b=a^p, a$
Similarly for prime $p$
Now if $p,q$ both divides $a^{pq}-a^q-a^p+a,$ the later must be divisible by lcm$(p,q)$
$\begin{align} {\bf Lemma}\ \ {\rm Suppose }\ \ &\forall a\!:\ \color{#c00}{P}(a)\equiv a\!\!\!\pmod{\! \color{#c00}p}\quad\![\text{OP is }\:\! P(a) = a^p,\, Q(a) = a^q]\\[,2em]
{\rm and}\ \ &\forall a\!:\ \color{#0a0}Q(a)\equiv a\!\!\!\pmod{\! \color{#0a0}q},\,\ \text{for some polynomials }P,Q\in\Bbb Z[x]\end{align}$
$\text{Then}\,\ \bmod pq\!:_{\phantom{|^{|^.}}}\!\! R(a) :=\color{#c00}P(\color{#0a0}Q(a))-\color{#c00}P(a)-\color{#0a0}Q(a)+a\equiv 0^{\phantom{|^{|^|}}}\!\!\!,\,$ if $\ (p,q)\!=\!1$.
${\bf Proof }\ \ {\rm mod}\ \color{#c00}p\!:\,\ R(a) \ \ \equiv\ \ \ \, Q(a)\ \: -\quad a\ \, - Q(a)+a\equiv 0^{\phantom{|^|}}\!\!\ \,$ (by erasing $\,\color{#c00}{P}\:\!$'s)
$\ \ \ \:\!{\rm and\,\ \ \ mod}\ \color{#0a0}q\!:\:\ R(a)\ \ \equiv \,\ \ \ P(a) \ -\, P(a)\ \ \:\! -\, \ \,a\ \, + a\equiv 0^{\phantom{|^|}}\!\!\ \,$ (by erasing $\,\color{#0a0}{Q}\:\!$'s)
so $\ p,q\mid R(a)\,\Rightarrow\, pq = {\rm lcm}(p,q)\mid R(a)^{\phantom{|^|}}\!\!$ by CCRT. $\,\ \bf\small QED$
Or $\ \{PQ(a),\ a\}\equiv \{Q(a), P(a)\}\,$ mod $\,p\,$ & $\,q\,$ so also $\!\bmod pq,\,$ so their sums are $\:\!\equiv\, \bmod pq.\,$ This view brings to the fore innate symmetry - see my other answer here.
Hint $\,\ \{a^{pq},a\} \equiv \{a^p,a^q\}\,$ both $\!\bmod p\,$ & $\bmod q,\,$ by little Fermat.
Hence $\, a^{pq}\!+\!a \ \equiv\ a^p\! + a^q\,$ mod $\,p,q,\,$ so also mod $\,pq = {\rm lcm}(p,q),\,$ by CCRT,
since addition $\,f(x,y)\, =\, x\! + \!y\,$ is $\rm\color{#c00}{symmetric}$ $f(\,x,y)= f(y,x),\,$ therefore its value depends only upon the (multi-)set $\{x,\,y\}.\,$
Generally if a polynomial $\,f\in\Bbb Z[x,y]\,$ is $\rm\color{#c00}{symmetric}$ then
$\qquad\qquad\quad\!\! \{A, B\} \equiv \{a,b\}\,\ {\rm mod}\,\ m\ \&\ n\ \Rightarrow\ f(A,B)\equiv f(a,b)\, \pmod{\!{\rm lcm}(m,n)}\qquad\quad$
a generalization of CCRT= constant-case optimization of CRT = Chinese Remainder, combined with a generalization of the Polynomial Congruence Rule to (symmetric) bivariate polynomials.
Another way.
Due to Little Fermat's theorem, $$a^{(p-1)(q-1)}-1\equiv 0\pmod{pq}\tag1$$ $$(a^{p-1}-1)(a^{q-1}-1)\equiv0\pmod{pq}\tag2$$ Now, $a^{p+q-1}(1)+a(2)$ results in $$a^{pq}-a^p-a^q+a\equiv0\pmod{pq}.$$
