Just Imagine

As known, $i^2=-1$ .

$i^{2016}=(i^{2})^{1008}=(-1)^{1008}=((-1)^2)^{504}=1^{504}=1$

Basically for $i^n =1$ , $n$ has to be a multiple of 4 because $1 = (-1)^2 = (i^2)^2 = i^4$ . Evidently, 2016 is divisible by 4 so $i^{2016} = 1$ .

We know that $i^4 = 1$ and $2016 \equiv 0$ ( $\text{mod 4}$ ) because according to the divisibility rules for 4 for $2016$ , $2016$ is divisible by $4$ $(\dfrac{16}{4}$ = integer ) so $i^{2016} = \boxed{1}$ )

$2016 \equiv 0 \left(\text{mod 2016}\right)$ .
And $i$ raised to any such number yields $\color{#69047E}{\boxed{1}}$ .

It helps to know the recurring pattern.

i^1 = sqrt(-1) i^2= -1 i^3 = -sqrt(-1) i^4 = (i^2)^2 = (-1)^2 = 1 i^5 = i^4*i^1 = 1 * i = i = sqrt(-1) i^6 = i^4 * i^2 = -1

....

It is a cycle with 4 steps. Knowing that it is 4 steps per cycle means that there are 504 cycles happening to arrive at this answer of 1, but they all cancel each other out when you work out the math. i^n only has 4 possible outputs for positive integers.