2016 is absolutely awesome 7

Use the factoring identity:

$a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2-...-ab^{n-2}+b^{n-1})$

The equation is $2016^n+1^n$

Substitute the corresponding values of $a$ and $b$ into the identity to yield $2017$ as the first factor and its prime too, so that's the smallest integer.

For E = {0, 2, 4, 6, 8, 10, ... to $\infty$ },

1
2
3
4
5
6
7
8
9

0   1
2   3
4   205
6   39991
8   14913081
10  9090909091
12  8230246567621
14  10371206370520815
16  17361641481138401521

This means $\displaystyle \frac{E^{E + 1} + 1}{E + 1}$ always an integer such that $E^{E + 1} + 1$ is always divisible by $(E + 1).$

Here, E = 2016 and therefore the expression is divisible by E + 1 = 2017. An expression added by 1 is not easily divisible by integers. Therefore, I were to guess that 2017 is the minimum possible value as answer, as I tried one by with integers from 2 to over 130 and spot checked to 170 but none is found!

Answer: $\boxed{2017}$