Why is $y^{x-1}-1$ divisible by $x$?

Question

I wanted to know if there is a way to prove that $y^{x-1}-1$ is divisible by $x$. Where $x$ is a prime number and is not equal to $y$, and $y$ is any positive whole number besides $1$. For example, $y=2$ and $x=13$ so $2^{13-1}-1$ is simplified to $4095$ which is divisible by $13$.

Answer 1

If you know some group theory this follows from the fact that the order a group element divides the group order: view $y$ as an element of the multiplicative group of units $(\mathbb{Z}/x\mathbb{Z})^*$. Since $x$ is prime, the latter has order $x-1$, whence $y^{x-1}\equiv 1$ mod $x$.

Answer 2

y^(x-1)-1=t y^x-y=yt By Fermat’s little theorem x|(y^x-y) and therefore x]yt Since (x,y)=1, x|t. If y has the factor x , t is not divisible by x, which is obvious. I also must say , this is Fermat’s little theorem. P.Ranawaka