It may be of interest to some that the claim in question may be proved by a simple application of double induction where we induct on two variables simultaneously. It is definitely overkill, but I think this proof technique is pretty sexy when it can be applied, and this is certainly one instance. Since using double induction is not too common, I thought I would include a blurb about it from David Gunderson's Handbook of Mathematical Induction before giving the proof.
Blurb: Many mathematical statements involve two (or more) variables, each of which vary independently over, say, $\mathbb{N}$. Some such statements can be proved by induction in different ways. Let $S(m,n)$ be a statement involving two positive integer variables $m$ and $n$. One method to prove $S(m,n)$ for all $m\geq 1$ and $n\geq 1$ is to first prove $S(1,1)$, then by induction prove $S(m,1)$ for each $m$, and then for each fixed $m_0$, inductively prove $S(m_0,n)$ for all $n$. [ p. 44]
Problem: Show that $S(m,n)$ is true where
$$
S(m,n) : (\forall m\in\mathbb{Z^+}\setminus\{1\})(\forall n\in\mathbb{Z^+}\cup\{0\})[(m-1)\mid (m^n-1)].
$$
Proof. First we prove that $S(m,0)$ is true for all $m\geq 2$.
Base step: The statement $S(2,0)$ is true since $(2-1)\mid (2^0-1)$.
Inductive step (inducting on $m$): For some $k\geq 2$, assume that $S(k,0)$ is true; that is, $(k-1)\mid (k^0-1)$. Since $((k+1)-1)\mid ((k+1)^0-1)$ is clearly true for all $k\geq 2$, we can see that $S(k,0)\to S(k+1,0)$, and by mathematical induction, for all $m\geq 2, S(m,0)$ is true.
Now fix an arbitrary $m_0$. Then $S(m_0,0)$ is true and so this is a base step for proving that for all $n, S(m_0,n)$ holds.
Inductive step (inducting on $n$): This step is of the form $S(m_0,\ell)\to S(m_0,\ell+1)$. Let $\ell\geq 0$ be fixed and assume that $S(m_0,\ell)$ is true, that is, $(m_0-1)\mid (m_0^\ell-1)$. Starting with the right-hand side of $S(m_0,\ell+1)$,
\begin{align}
m_0^{\ell+1}-1 &=m_0(m_0^\ell-1)+(m_0-1)\\[0.5em]
&=m_0[\eta(m_0-1)]+(m_0-1)\tag{by $S(m_0,\ell);\eta\in\Bbb{Z}$}\\[0.5em]
&=(\eta m_0+1)(m_0-1)\tag{factor}\\[0.5em]
&=\gamma(m_0-1),\tag{$\gamma=\eta m_0+1$}
\end{align}
which is a multiple of the left-hand side of $S(m_0,\ell+1)$, where $\gamma=\eta m_0+1$ and $\gamma\in\Bbb{Z}$ since $1,\eta,m_0\in\Bbb{Z}$ (closure of addition and multiplication in $\Bbb{Z}$). Since $m_0^{\ell+1}-1=\gamma(m_0-1)$, we can see that, by definition, $(m_0-1)\mid (m_0^{\ell+1}-1)$; hence, by induction, for each fixed $m_0$ and $n\geq 0, S(m_0,n)$ is true, completing the inductive step.
Since $m_0$ was arbitrary, by induction, for all $m\geq 2$ and $n\geq 0$ the statement $S(m,n)$ is proved. $\Box$
