Every proper maximal subgroup of a $p$-group $P$ is normal and has index $p$.
I tried to search online by I can't get a complete proof.
Take $M$ to be maximal and $Z$ to be central subgroup of order $p$.
Case 1: $Z\subset M$
We prove by induction on $|P|$.
For the base case, $|P|=p$.
There is no maximal subgroup in $P$, hence the result holds
Assume that the result holds for all $p$-group $Q$ with $|Q|<|P|$
Note that $|Z|>1$. Hence $|P/Z|<|P|$.
Suppose that $M/Z$ is not a maximal subgroup of $P/Z$.
Then there exists a subgroup $H$ such that $M/Z\subset H/Z \subset P/Z$.
But this also implies that $M\subset H \subset P$.
This means that $M$ is not maximal, a contradiction.
Hence $M/Z$ is a maximal subgroup of $P/Z$.
By induction hypothesis, $M/Z$ has index $p$ and $M/Z\trianglelefteq P/Z$.
First, we can conclude that $M \trianglelefteq P$.
Second, note that $[P/Z:M/Z]=p$. Hence $[P:M]=p$.
Case 2: $Z\not\subset M$
This implies that $P=ZM$.
Let $p=zm\in P$ where $z\in Z$ and $m\in M$.
Clearly, $p^{-1}Mp=M$. So we see that $M \trianglelefteq P$.
But for this case, I have no idea on how to prove $M$ has index $p$.
Best Answer
Theorem 1: Every finite $p$-group is nilpotent.
Proof: see here.
Theorem 2: A finite group is nilpotent if and only if all its maximal proper subgroup are normal.
Proof: see here.
Theorem 3: A proper normal maximal subgroup of a finite group has prime index.
Proof: see here.
Corollary: Every proper maximal subgroup of a $p$-group is normal and has index $p$.