Fermat's little theorem question: why isn't $a^p \equiv 1$? Fermat's little theorem says that $a^p \equiv a \pmod p$.
I have kind of a stupid question. Since $p \equiv 0\pmod p $, why isn't $a^p \equiv a^0 \equiv 1 \pmod p$ ?

You can't take the modulo of the exponent like that. Think of $a^p$ as $\underbrace{a*a*\dots*a}_{p\text{ times}}$. If it was $p*a$ then what you did would be justified, but not with exponents. A full proof of Fermat's little theorem is off-topic, but this should answer your question!
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Edit to answer OP's comment on the original post:
When in doubt you can always go back what $a = b \text{ (mod n)}$ means.
$$
\begin{align}
a = b \text{ (mod n)} \text{ is equivalent to } a = b + k * n \text{ for some } k \in \mathbb{Z}
\end{align}
$$
From that it is easy to show that $n * a = 0 \text{ (mod n)}$:
$$
\begin{align}
n * a = b + k * n &\implies 0 = b + (k + a) * n\\
&\implies 0 = b \text{ (mod n)}
\end{align}
$$
But you can't do the same with $a^n$.
