Euler-Fermat Theorem So I am trying to teach myself number theory, and while trying to work on some exercises I got stuck trying to prove that, for all $n \in \mathbb{Z}$, 
$$
n^{91} \equiv n^{7} \bmod 91
$$
What I first thought of was applying the theorem directly and then multiply by the number of n's that was necessary, getting something like
\begin{array}{ccc}
n^{72} &\equiv& 1 \bmod 91 \\
n^{91} &\equiv& n^{19} \bmod 91
\end{array}
Which is the same as finding 
\begin{array}{ccc}
n^{19} &\equiv& n^{7} \bmod 91 \\
n^{12} &\equiv& 1\ \bmod 91
\end{array}
But then I got stuck. Does anyone have anything that can give me a push? I am not looking for the answer, but more of a hint. Thanks in advance!

Hint: $91 = 13 \times 7$, so $a \equiv b \pmod {91}$ iff  $a \equiv b \pmod {13}$  and  $a \equiv b \pmod 7$. See CRT
