Prove that $τ(m^n)$ and $n$ are coprime $(m,n ∈ N^+)$

Question

We have that $τ(n)=\sum_{d|n} 1$ is the number of dividers of n.

Dividers of $m^n$ are $1,m,m^2,...,m^n$ then we have that $τ(m^n)=n+1$.

Is this correct so far?

Now we must prove that $τ(m^n)$ and $n$ are coprime, and I've though of using Bezouts theorem. Is this also a correct way?

Answer 1

Hint

If a prime $p$ divides $n$ and $n+1$ then it divides $(n+1)-n = 1$ as well.

Answer 2

Hint: Remark that if $m=\Pi p_i^{l_i}$, where $p_i$ is prime and $p_i$ and $p_j$ are coprime if $i\neq j$, $r(m)=\Pi(l_i+1)$ this implies that $r(m^n)= \Pi(nl_i+1)$.