Okay so I tried to solve it on my own but I have no idea if it is correct. Here's what i think:
Since |G| = p^m x....x p^n . Then there exists a subgroup p where p^r/p = p^r-1 . I don't know how to show it is normal and i also think we should use induction but I don't know how?
a group of such order cannot have a trivial center. Why? Use the class equation and write
$G=Z(G)+\sum\limits_{i=1}^r \frac{G}{C_g(G)}$ where $g_1,g_2\dots g_r$ are representatives for the non-trivial conjugacy classes.
If the center where trivial it would mean one of the classes say the class of $g$ is not a multiple of $p$ which in turn would imply it's centralizer is all of $G$ which would mean that $g$ is in the center, contradicting the center is trivial.
Since the center is not trivial it has a subgroup $H$ of order $p$, since any subgroup of the center is normal we can take $\frac{G}{H}$ which has order $p^{\alpha-1}$ and apply the inductive hypothesis to obtain a normal subgroup of $\frac{G}{H}$ of order $p^{\alpha-2}$ which we shall call $E$
The preimage or pullback of $E$ under the natural homomorphism from $G$ to $\frac{G}{H}$ will be a normal subgroup of $G$ of order $p^{\alpha-1}$.
Why? The preimage of a normal subgroup under any homomorphism is always normal.
Moreover let $I$ be the preimage of $E$. consider the restriction of the natural homomorphism from $I$ into $E$, this map is surjective. So by the first ismorphism theorem we have $\frac{I}{H}\cong E$ looking only at the orders we get $|E|=p^{\alpha-2}p=p^{\alpha-1}$
So $I$ is indeed a normal subgroup of $G$ of the desired order.
Hints to proof by induction: let $\;|G|=p^r\;,\;\;p\;$ a prime:
== $\;|Z(G)|>1\;$ (if you need to prove this upmost important fact of finite $\;p$ - groups use the Class Equation
== Take $\;1\neq z\in Z(G)\;$ of order $\;p\;$ and look at $\;K:= G/\langle z\rangle\;$ . Since $\;|K|=p^{r-1}\;$ the inductive hypothesis gives us a normal subgroup $\; \overline H\le K\;$ of order $\;p^{r-2}\;$.
== Use now the Correspondence Theorem to deduce that $\;\overline K=K/\langle z\rangle\;$ , with $\;\langle z\rangle\le K\le G\;$ , and $\;K\lhd G\;,\;\;|K|=p^{r-1}\;$
