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

Proof

 * $\left\lvert{G}\right\rvert = k p^n: p \nmid k$
 * $\left\lvert{G}\right\rvert = k p^n: p \nmid k$

where $\left\lvert{G}\right\rvert$ denotes the order of $G$.

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

From the First Sylow Theorem:
 * $\exists H \le G: \left\lvert{H}\right\rvert = k$

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

Thus:
 * $\exists h \in H: \left\lvert{h}\right\rvert \mathrel \backslash k \implies \left\lvert{h}\right\lvert \nmid p$

where $\backslash$ denotes divisibility.

Thus:
 * $\exists h \in G: \left\lvert{h}\right\rvert \ne p^n: n \in \Z$

Thus by Proof by Contradiction, $\left\lvert{G}\right\rvert$ must be a power of $p$.