Prove the order gH must divide order of g....help

Let G be a finite group and let H be a normal subgroup of G. Prove
that the order of the element gH in G/H must divide the order
of g in G.

A hint in the book said order of g = n and to suppose (gH)^n = (g^n)H = eH = H

Does anyone know how to finish this?

By definition of order of gH, it is the smallest positive integer m such that (gH)^m=H
Dividing m by n we have m = pn + r where r (gH)^(pn+r) = [ (gH)^n ]^p (gH)^r = H
But, from the hint given, [ (gH)^n ]^p = H^p = H and therefore (gH)^r = H. But we had said above that m was the smallest positive integer for which such equality was possible so r=0 and therefore m =pn , i.e. m divides n