Base $n$ ?

The number listed is equivalent to:

$(n-5) + (n-4) \cdot n + (n-3) \cdot n^2 + (n-2) \cdot n^3 +(n-1) \cdot n^4$

Consider the number under $\mod {(n^2-1)}$ . $n = n \mod{(n^2-1)}$ and $n^2 = 1 \mod{(n^2-1)}$ , a fact that follows naturally by rewriting $n$ as $n-1+1$ and dividing by $n-1$ . Using these two facts, it is easily proven by multiplication that $n^a = 1 \mod{(n^2-1)}$ when $a$ is even, and $n^a = n \mod{(n^2-1)}$ when $a$ is odd.

$(n-5) + (n-4) \cdot n + (n-3) \cdot n^2 + (n-2) \cdot n^3 +(n-1) \cdot n^4 \mod{(n^2-1)}$

$(n-5) + (n-4) \cdot n + (n-3) + (n-2) \cdot n +(n-1) \mod{(n^2-1)}$

$2n^2-3n-9 \mod{(n^2-1)}$

$2-3n-9 \mod{(n^2-1)}$

$-3n-7 \mod{(n^2-1)}$

This is not $0 \mod {(n^2-1)}$ , so the answer is $False$

One counter example: In base 10, 98765 isn't divisible by 99, since it doesn't divide by either 11 or 9. Therefore, the answer is $\boxed{false}$ .