As $2017$ is prime, using Wilson's Theorem $$2016!\equiv-1\pmod{2017}$$
$$\iff 2015!\cdot 2016\equiv-1\pmod{2017}$$
$$\iff 2015!\cdot (-1)\equiv-1\pmod{2017}\text{ as }2016\equiv-1\pmod{2017}$$
So, $\cdots$
As $\displaystyle2017\equiv17\pmod{1000},2017^m\equiv17^m$ for any integer $m$
Use Carmichael function, $\lambda(1000)=100$
$$\implies17^{(2016^{2015})}\equiv17^{(2016^{2015})\pmod{100}}\pmod{1000}$$
Now $2016\equiv16\implies2016^{2015}\equiv16^{2015}\pmod{100}$
As $(16,100)=4$ let use find $16^{2015-1}\pmod{100/4}$
$16^{2014}=(2^4)^{2014}=2^{8056}$
As $\displaystyle\lambda(25)=\phi(25)=20$ and $8056\equiv16\pmod{20},2^{8056}\equiv2^{16}\pmod{25}$
$2^8=256\equiv6\pmod{25}\implies2^{16}\equiv6^2\equiv11$
$\implies16^{2014}\equiv11\pmod{25}$
$\implies16^{2014+1}\equiv11\cdot16\pmod{25\cdot16}\equiv176\pmod{400}\equiv76\pmod{100}$
$$\implies17^{(2016^{2015})}\equiv17^{76}\pmod{1000}$$
Now $$17^{76}=(290-1)^{38}=(1-290)^{38}\equiv1-\binom{38}1290+\binom{38}2290^2\pmod{1000}$$
Now $\displaystyle38\cdot29=(40-2)(30-1)\equiv2\pmod{100}\implies\binom{38}1290\equiv20\pmod{1000}$
and $\displaystyle\binom{38}229^2=\dfrac{38\cdot37}2(30-1)^2\equiv3\pmod{10}$
$\displaystyle\implies\binom{38}2290^2\equiv3\cdot100\pmod{10\cdot100}$
Best Answer
let $$ n = 2015^{2016}-2017 $$ If you reduce everything modulo $100$ you'll get $$n \equiv 15 ^ {2016} - 17 \,\, \text{(mod 100)}$$
To reduce $15^{2016}$ , we calculate $$15^2=125\equiv 25 \,\,\text{(mod $100$)}$$ so $$15^{2016} = 25^{1008}$$ As $25^2 = 625 \equiv 25 \,\,\text{(mod 100)}$, every power of $25$ is equivalent to $25$ (mod 100).
In the end:
$$ \begin{align} n &\equiv 15 ^ {2016} - 17 \,\,&\text{(mod 100)} \\ &\equiv (15^2)^{1008} - 17 \,\,&\text{(mod 100)}\\ &\equiv 25^{1008} - 17 \,\,&\text{(mod 100)}\\ &\equiv 25 -17 \,\,&\text{(mod 100)}\\ &\equiv 8 \,\,&\text{(mod 100)} \end{align} $$
So the last two digits are $08$.