modular arithmetic - Prove if $a^{(p)(p-1)} equiv 1$ (mod $p$), then $p$ divides $k$, where $k$ is an integer such that $a^{p(p-1)} - 1 = pk$, and $p$ is a prime number.

I am trying to prove that $a^{p(p-1)} \equiv 1$ (mod $p^2$). I have reduced the equation to essentially solving the the problem above, but I am not sure how to proceed. I've tried to use Fermat's Little Theorem and the fact that $a^{p(p-1)} \equiv 1$ (mod $p$) and that $a^{(p-1)} \equiv 1$ (mod $p$), but I haven't gotten anything concrete.

Answer

Write $a^{p-1}=1+pt$ with $t \in \mathbb Z$.
Then
$$a^{p(p-1)} = (1+pt)^p = \sum_{i=0}^p \binom{p}{i}(pt)^i \equiv 1+(pt)^p \equiv 1 \bmod p^2$$

because $\displaystyle\binom{p}{i}$ is a multiple of $p$ for $0 < i < p$.