Prove any subgroup (H) of a cyclic group (G) is itself cyclic

Approach:

Let

Assume H is non-trivial

There is a such that k is the smallest +ve power among all that are in H. Well order principle of natural numbers ensures there will always be a unique such k as long as H is non-trivial

Claim

Let

where

obviously (closure property of H)

Thus r can't be a +ve number (as k is the smallest). Thus r=0.

Hence

And, (By definition)

Is this proof 'rigorous' enough?