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).
A: Nice observation! You noticed $\!\bmod 100\,$ the powers of $\,17\,$ have the following structure
$$\begin{array}{r|r r}
n & 17^{\large n}\!\! & \\
\hline
0 &\ \  01 & 17 & 89 & 13\\
4& 21 & 57 & 69 & 73\\
8& 41 & 97 & 49 & 33\\
12& 61 & 37 & 29 & 93\\
16& 81 & 77 & 09 & 53\\
\end{array}\qquad$$
This structure is due to $\,17^{\large 4}\equiv 21\pmod{\!100}$ so multiplying by it has the following effect
$$\bmod 100\!:\ \ 21(10t+u) = 210t+21u\equiv 10(t+\color{#c00}{2u}) + u\quad\ $$
thus it has the same units digit $\,u\,$ and the tens digit $\,t\,$ is incremented by $\,\color{#c00}{2u}\bmod 10,\,$ e.g. if units digit $\,u = 7\ $ ($\rm 2nd$ column) $ $ the tens digit $\,t\,$ is incremented by $\,2\cdot 7\bmod 10 = 4$
We don't need the whole table:  $\,17^{\large 4}\equiv 21\pmod{\!100},\,$ &  $\rm\color{#0a0}{BT}$ = Binomial Theorem imply
$$\bmod 100\!:\,\ 17^{\large 2016}\equiv (17^{\large 4})^{\large 504}\equiv (1\!+\!20)^{\large 504}\overset{\rm\large\color{#0a0}{BT}}\equiv 1+ 504(20)\equiv 1+ 4(20)\equiv 81\qquad$$
by $\,20^{\large k}\equiv 0\,$ for $\,k\ge 2.\,$ Hence we conclude $\, 17^{\large 2017}\equiv 17(81)\equiv 17(-19)\equiv -23\equiv 77$
Alternatively $\ 17^{2016}\equiv (17^{\large 2})^{\large 1008}\equiv (-1\!+\!90)^{\large 1008}\equiv 1\!-\!1008(90)\equiv 1\!-\!20\equiv 81,\,$ but I chose the above to stay closer to your observation.
The use of the Binomial Theoremto lift $\,17^{\large 4}\equiv 1\pmod{\!10}\,$ up to modulo $10^{\large 2}$ is a special case of more general methods, e.g. see LTE = Lifting The Exponent here and also this  result
$$a\equiv b\!\!\! \pmod{\!kn}\,\Rightarrow\,a^{\large k}\equiv b^{\large k}\!\!\!\! \pmod{\!k^2n}$$
A: It suffices to evaluate $ (2017)^{2017} \pmod {100}: $
$$ \equiv (17)^{2017} \equiv (17^{2})(17) \equiv (-11)^{1008}(17) $$
$$ \equiv (21)^{504}(17) \equiv (21^{5})^{100}(21^{4})(17) \equiv (1)^{100}(21^{4})(17)$$
$$ 77 \pmod {100}$$
Therefore, the last two digits are $77$.
