An algebra problem by Alexander Koran

We note that $i^k = \begin{cases} 1 & \text{if } k \text{ mod }4 = 0 \\ i & \text{if } k \text{ mod }4 = 1 \\ -1 & \text{if } k \text{ mod }4 = 2 \\ -i & \text{if } k \text{ mod }4 = 3 \end{cases}$

Therefore, $\displaystyle P = \prod_{n=1}^{2017} i^{n^n+n!} = \prod_{n=1}^{2017} i^{(n^n+n!) \text{ mod }4}$

We note also that when $n$ is even $n^2 \equiv (2m)^{2m} \equiv 0 \text{ (mod 4)}$ and when $n$ is odd $n^2 \equiv (2m+1)^{2m+1} \equiv 1 \text{ (mod 4)}$ . Also that $n! \equiv 0 \text{ (mod 4)}$ for $n \ge 4$ .

Therefore, we have

$\begin{aligned} P & = \underbrace{i^{1+1}}_{n=1} \cdot \underbrace{i^{0+2}}_{n=2} \cdot \underbrace{i^{1+2}}_{n=3} \cdot \prod_{n=4}^{2017} i^{(n^n+n!) \text{ mod }4} \\ & = i^2 \cdot i^2 \cdot i^3 \cdot \underbrace{i^{0\times 1007}}_{even} \cdot \underbrace{i^{1 \times 1007}}_{odd} \\ & = i^{1014} = i^{1014 \text{ mod }4} = i^2 = \boxed{-1} \end{aligned}$

2017 ^2017 will end with 7 using c clicity ans 2017! Will end with zero then u can solve it using iota to the power 07 leaves remainder 3 so the answer is -1

This question may be solved in two ways.

(1) Solve by finding remainder by dividing by 4

(2) In the answer field u allows only integer answers.So $i^m=i,-1,-i,1$ .So there are only integers are possible .Anyone can solve within maximum two tries.

So , I suggest you to use multiple correct options.