What is $[GF(p^n):GF(p^m)]$ if $m|n$?

Question

I am tempted to use the formula \begin{equation} [GF(p^n):GF(p)] = [GF(p^n):GF(p^m)][GF(p^m):GF(p)] \end{equation} \begin{equation} n = [GF(p^n):GF(p^m)]m \end{equation} and thus \begin{equation} [GF(p^n):GF(p^m)] = \frac{n}{m} \end{equation} But I suspect this is too simple to be a proof of this.