Q:

What are the last two digits of $2017^{2017}$?

What are the last 2 digits of $2017^{2017}$?
Notice that 
$$2017 (2017) = 2017 ( 2000 + 10 + 7) = (....000) + (....70) + (2017 \times 7)$$
so the last two digits of $2017^{2}$ are the last two digits of $70$ + last two digits of $17 \times 7$. These are $89$. 
Similarly then 
$$ (2017^2) (2017) = (....89)(2000 + 10 + 7) = (....000) + (....90) + (....89)\times7 $$
So the last two digits of $2017^{3}$ are the last two digits of $90$ + last two digits of $89 \times 7$. These are $13$.
for $2017^{4}$, the last two digits are the last two of $30$ + last two of $13 \times 7$.

But somehow, I have not find the period for the last two digits of $2017^{n}$. I have tried until $n=36$, still no period.
The answer should be one of $77,81,93,37,57$. 

I just found a pattern: 
$$ n=1,n=2,n=3,n=4 \rightarrow 17, 89, 13,21 $$
this pattern repeats as: the 1st one with difference 40, 2nd with diff 80, 3rd with diff 60, and 4th with diff 20. 
$$ n=5,n=6,n=7,n=8 \rightarrow 57, 69, 73,41 $$
and so on. The overall period begin again from $n=17$. 
So the last digit is 77. The math explanation?

A:

As we're interested in the last two digits, it suffices to compute this modulo $100$. Hence
$$
 2017 ^{2017} \equiv 17^{17} \equiv 77 \pmod{100}.
$$
The first equivalence holds by Euler's theorem (see the mothertopic mentioned by J. Lahtonen).

