Products, sums and divisors 2

The only value for $x$ is $\boxed{0}$ .

Let $N = \frac{x^{n-1}}{n}$ , then when $x$ is 0:

$N = \frac{0}{n}$

Since for all $n$ , $n | 0$ , so we have the only value for $x$

Now it is by chance got the right answer but I actually have many reasons to believe its not; here's some which is proved by using some code:

If you look at the console at the right, you can already see two solutions why and the code and computer combined will keep going to kind as many solutions to this formula as possible; so the answer really is infinite.

I can prove this because you can still add zeroes to 6 or 8 like 6,60,600,etc or 8,80,800,etc and still keep the power 'n' the same (which is 2 in this case) and make an infinite amount of solutions

The answer must be $x=1$ , for all $n$ $\epsilon$ $\Z^+$ , $x^n$ is divisible by $nx$ , i.e $1^5$ is divisible by $5*1$ .