Proving Fermat's Little Theorem with algebra

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(a) Let $p$ be a prime and let b be a non zero element of the field $\mathbb{Z}_{p}$. Show that $b^{p-1} = 1$. Hint Lagrange

My attempt

$|\mathbb{Z}_{p} \setminus \{0\}| = p-1$

since $b \in \mathbb{Z}_{p} \setminus \{0\}$

$\langle b\rangle \leq \mathbb{Z}_{p} \setminus \{0\}$

From Lagrange $\langle b\rangle | \mathbb{Z}_{p} \setminus \{0\}$

suppose $\langle b \rangle = n$

then $\implies n \mid p-1$

$\implies p-1 = nk$

$\implies b^{p-1}=(b^{n})^{k} = e^{k} = e = 1$

(b) Use (a) to prove that if $p$ is prime and a is an integer the p divides $a^{p}-a$.

And now I have to use the above fact to prove Fermat's Little Theorem.. any hints?