Let G be a group of order p^m where p is prime number and m is a positive integer. Show that G contains an element of order p.

Let G be a group of order ${\displaystyle p^m}$, where p is a prime number and m is a positive integer.
Then,
${\displaystyle \left|G\right|=p^m=p{p}^{m-1}=pn,}$
Where ${\displaystyle n=p^{m-1}}$
Then, G is a finite group and the prime p divides the order of G, i.e.,
Proving the result by induction on n,
Let n=1
Then, ${\displaystyle p^{m-1}=1}$ and hence p=1.
Then, |G|=1
Thus, the group has only one element of order 1. The result holds for n=1
Next, suppose every group with pk elements contains an element of order p for every k Consider the class equation of G, ${\displaystyle \left|G\right|=\left|Z\left(G\right)\right|+\sum _{a\ne Z\left(G\right)[G:C\left(a\right)]}}$
If G=Z(G), then G is abelian and hence the result follows.
Suppose ${\displaystyle G\notin Z\left(G\right)}$
Then, for every a !in Z(G),
${\displaystyle C\left(a\right)\subset G}$, and hence ${\displaystyle \left|C\left(a\right)\right|<G}$.
If for some ${\displaystyle a\notin Z\left(G\right),p\left|C\left(a\right)\right|}$ then, by the induction hypothesis, there is an element a of order ${\displaystyle p\in C\left(a\right)}$ and hence ${\displaystyle \in G}$.
Otherwise, for every ${\displaystyle a\notin Z\left(G\right),p\mid \left|C\left(a\right)\right|.}$
Then, ${\displaystyle p\mid \sum _{a\ne Z\left(G\right)[G:C\left(a\right)]\mid }}$ and the class equation implies that ${\displaystyle p\left|Z\left(G\right)\right|.}$
Thus, it follows that Z(G) and hence G contains an element of order p.