How to prove this using Fermat's Little Theorem?

Question

Suppose that $p$ and $q$ are distinct odd primes and $a$ is an integer such that $\text{gcd}(a,pq)=1$. Prove that $a^{(p-1)(q-1)+1}\equiv a \pmod{pq}$.

Answer 1

Hint: $(p-1)(q-1)$ is the value of Euler's phi-function; then go from there.

Answer 2

Use Fermat's Little Theorem to say $a^{p-1}\equiv1 \mod p$ and $a^{q-1}\equiv 1 \mod q.$

Therefore $a^{(p-1)(q-1)}\equiv 1 \mod p,q.$

Therefore $a^{(p-1)(q-1)}\equiv 1 \mod pq,$ using the Chinese Remainder Theorem.

Now multiply both sides by $a$.