Any divisor of $(n + 1)^p - n^p$ is $1 \textrm{ mod } p$ I'm trying to work my way through the following problem.

Let $n$ be a natural number, $p$ a prime, and $d$ a divisor of $(n + 1)^p - n^p$.
Show that $d \equiv 1 \textrm{ (mod } p \textrm{)}$.

I'm not sure where to start on this. I can see $(n + 1)^p - n^p \equiv 1 \textrm{ (mod } p)$ and ofcourse $(n + 1)^p - n^p \equiv 0 \textrm{ (mod } d)$. So definitely $p, d$ are coprime. Also $(n + 1)^p - n^p \equiv 1 \textrm{ (mod } n)$. This feels like it shouldn't be too hard a problem but I'm not entirely sure what I'm missing.
Let $q$ be a prime divisor of $(n+1)^p-n^p$.

Shall show $q\equiv 1\ mod \ p$

Clearly $(q,n)=(q,n+1)=1$ and hence we have in the field $\mathbb Z_q$
$$(n+1)^p=n^p \ in  \ \ \mathbb Z_q$$
$$\implies ((n+1)n^{-1})^p=1 \ in \ \mathbb Z_q$$

SO elementary group theory says $order \  ((n+1)n^{-1}))=p \ in \ \ \mathbb Z_q^*$ 

So $p|q-1$ and hence any divisor $\equiv 1 \ mod \ p$
Well, the binomial theorem gives
$$(n+1)^p - n^p = \sum_{i=0}^p {p\choose i} n^i - n^p = \sum_{i=0}^{p-1}{p\choose i} n^i.$$
But $p$ divides ${p\choose i}$ for each $i=1,\ldots,p-1$ and so taken modulo $p$ there is only one summand left $(i=0)$, which is $1$.