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!
 A: 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
A: Let's test for $$n^{p\cdot q}-n^p\pmod {p\cdot q}$$ where $n$ is any integer and $p,q$ are prime
Using Fermat's Little Theorem, $a^q\equiv a\pmod q$ for any integer $a$
$$n^{p\cdot q}-n^p=n^p\cdot\left((n^{q-1})^p-1^p\right)$$
$$ \text{ which is divisible by }n\cdot(n^{q-1}-1)=n^q-n \text{ which is divisible by } q$$ 
Again,
$$\text{if }p|n,p|(n^{p\cdot q}-n^p)$$
$$\text{else } (n,p)=1 \text{  and } n^p\equiv n\pmod p$$
$$\implies n^{p\cdot q}-n^p=(n^p)^q-(n^p)\equiv n^q-n\pmod p$$
$$p|(n^q-n)\iff p|(n^{q-1}-1)\text{ as }(n,p)=1$$
If we denote the order of $a\pmod p =\text{ord}_pa$ such that $(a,p)=1,$
as $p$ is prime, the highest value of $\text{ord}_pa$  will be $p-1$ and the other values of $\text{ord}_pa$ will divide $p-1$
Hence , we need $(p-1)$ to divide $(q-1)$ so that $p|(n^q-n)\implies p|(n^{p\cdot q}-n^p) $
Now, we know, $a\equiv b\pmod m$ and $a\equiv b\pmod n$
$\implies a\equiv b\pmod {[m,n]}$ where $[m,n]=$lcm$(m,n)$
If $n^{p\cdot q}\equiv n\pmod q$ and $n^{p\cdot q}\equiv n\pmod p$
$\implies n^{p\cdot q}\equiv n\pmod {[p,q]}$
Now, $[p,q]=p\cdot q$ as $(p,q)=1$ as $p,q$ are distinct primes
