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?
 A: Hint: Compute $4^1\pmod{13}, 4^2\pmod{13}, 4^3\pmod{13},\ldots$ and see if you can find a pattern.
A: Use lil' Fermat:
$4762\equiv 4\mod 13$, hence
$$4762^{5367}\equiv 4^{5367}\equiv4^{5367\bmod 12}=4^3=64\equiv -1\mod 13. $$
A: 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 $4^{5367}\equiv 4^3\equiv 64\equiv 12 \bmod 13$
A: Observe that $4^6 = 1 \pmod{13}.$ Thus, $4^{5364}=1 \pmod{13}.$
It follows that $$4^{5367}=4^3 \pmod{13}=-1\pmod{13}$$
A: HINT
Use Fermat's Little Theorem$$a^{p-1}\equiv 1 \pmod p$$
to
$$4^{5367} \pmod{13}$$
and consider that 
$$5367\equiv 3 \pmod {12}$$
A: Note that $$4^3=64\equiv -1\mod 13$$ $$5367=3\times 1789$$ $$4^{5367} =4^{3\times 1789}\equiv -1^{1789}=-1 mod (13).$$
