Order of Finite p-Group is Power of p/Proof 1

Proof

 * $\order G = k p^n: p \nmid k$
 * $\order G = k p^n: p \nmid k$

where $\order G$ denotes the order of $G$.

By Divisors of Power of Prime:
 * $k \nmid p^n$

From the First Sylow Theorem:
 * $\exists H \le G: \order H = k$

where $H \le G$ denotes that $H$ is a subgroup of $G$.

Thus:
 * $\exists h \in H: \order h \divides k \implies \order h \nmid p$

where $\divides$ denotes divisibility.

Thus:
 * $\exists h \in G: \order h \ne p^n: n \in \Z$

Thus by Proof by Contradiction, $\order G$ must be a power of $p$.