It is known that $p$ divides the binomial coefficient $\binom{p}{i}$ for $1\leq i\leq p-1$. So from the binomial theorem, it is not hard to see $ (a+1)^p\equiv a^p+1 $ modulo $p$.
Is there a way to derive Fermat's little theorem $a^p\equiv a$ mod $p$, without appealing to Lagrange's theorem? I feel like I could say $(\mathbb{Z}/p\mathbb{Z})^\times$ is a cyclic group of order $p-1$, thus $(a+1)^p\equiv a+1$, hence $a+1\equiv a^p+1$, but this seems to miss this point since I would know $a^p\equiv a$ from the get go.