Proof conclusion: Let m,n,s $\in$ $\mathbb{N} $\ $\{0\}$. Prove that $s| \tau(m^{s}) - \tau(n^{s})$, where $\tau$ is the divisor function.
Hello so my question is that, I've gotten to the point in the proof where I had used the divisor function definition and have $\tau(m^{s})$ = $\prod$ something +1 , and $\tau(n^{s})$ = $\prod$ something +1.
I know, that since the last terms of both polynomials are equal to 1. Somehow this results that $s| \tau(m^{s}) - \tau(n^{s})$.
But I don't understand why the +1 would indicate that: $s| \tau(m^{s}) - \tau(n^{s})$ holds.