In proving Sylow's Theorem (1), how could we start by assuming inductively that Sylow p-subgroups exist? In proving this Sylow's Theorem:

The proof starts by assuming inductively the existence of Sylow's p-subgroups as following:

Isn't it circular?
 A: This is the usual structure of a strong induction proof.  The intuitive idea is that if the statement fails for some set of numbers, there is a smallest one that it fails for.  If we show that (it is true for $n=1$ and) for all $n$, the fact that it is true for all numbers $1$ through $n$ implies it is true for $n+1$ shows there is no smallest number that it fails for, so it must be true for all $n$.  Here the induction is on the size of $G$.  The claim is that if there is a group without a Sylow subgroup, there is one of minimum order. When we consider that group, all smaller groups have Sylow subgroups.  We derive a contradiction and conclude that there is no smallest group without a Sylow subgroup.
A: Note that it first assumes the existence of Sylow subgroups for all groups of order less then $|G|$. By that assumption, it proves the existence of Sylow subgroup of $G$. Since $|G| = 1$ is a group of order less than any other groups, and $G$ is definitely a Sylow subgroup, it completes the proof.
