Why does p divide [G:H] in the proof of the first Sylow thm?

Question

Let G be a group of order $p^{n}m$ with p prime and (p,m)=1 and let H be a subgroup of G of order $p^{i}$, with $0<i<n$. Then p divides [G:H].

In every proof of Sylow thm this came out and was stated like it's so trivial. But it looks not so trivial to me, and i just got stuck here.

Any hints or comments will be appreciated.

Answer 1

$$[G:H] = |G|/|H| = p^n m / p^i = p^{n-i} m$$ is divisible by $p$ because $n>i$.