Find $4762^{5367} \pmod{13}$ 
Find $4762^{5367} \pmod{13}$

So I started off and got:
$$4762 = 4 \pmod{13}$$
Therefore, $$4762^{5367} = 4^{5367} \pmod{13}$$
But I did not know how to proceed after this.
Any help?

For a small modulus like $13$, it's worth just looking for the pattern that $4$ follows on exponentiation:
$\bmod 13: \\
\begin{align}
\qquad 
4^0 &\equiv 1\\
4^1 &\equiv 4\\
4^2 &\equiv 16 \equiv 3\\
4^3 &\equiv 12 \equiv -1\\
4^4 &\equiv -4 \equiv 9 \\
4^5 &\equiv -3 \equiv 10\\
4^6 &\equiv -1^2 \equiv 1\\
\end{align}$
And what we're after is closing the repeat, which we can be sure will happen because there are a limited number of values $(\phi(13)=12)$ that the values can take.
We could be sure in advance, in fact, that $4^6\equiv 1\bmod 13$ because Fermat's little theorem gives $a^{12}\equiv 1 \bmod 13$ (for $13\nmid a$) and thus  $2^{12}=4^6\equiv 1 \bmod 13$.
Then $4^{5364} = 4^{6\times 894} \equiv 1^{894}\equiv 1 \bmod 13$. So $4^{5367}\equiv 4^3\equiv 64\equiv 12 \bmod 13$
