Let $p$ be a prime number, and let $G$ be a $p$-group: $|G| = p^r$. Prove that $G$ contains a normal subgroup of order $p^k$ for every nonnegative $k \le r$.
The answers here and here use induction but they assume $G$, where $|G|=p^r$, has normal subgroups of order $p^k$ for $k <r$. Induction should start by assuming for every $p$-group of order $p^k$ where $0\le k <r$, there exists normal subgroups of order $p^i$ where $0 \le i \le k$.
We have to show there exists normal subgroups of order $p^i$ where $0 \le i \le r$.
If $|G|=p^0=1$, then vacuous. If $|G|=p$, then $\{1\}$ and $G$ are normal subgroups of order $p^0$ and $p^1$.
Suppose the statement is true for $p$-groups of order $p^k$ where $k < r$. Let $|G|=p^r$.
Since $G$ is a $p$-group, it has a non-trivial center, $Z(G)$. So, $Z(G)$ is a $p$-group. Since the center is abelian, then, by Cauchy's theorem, there exists an element of order $p$ and thus a subgroup of order $p$, say $N$. Since $N \subset Z(G)$, then $N$ is normal in $G$.
Consider $G/N$. Then $|G/N|=p^{r-1}$. By the induction hypothesis, there exists normal subgroups of order $p^i$ for $0\le i \le r-1$. By the correspondence theorem, these normal subgroups have the form $H_0/N, H_1/N, \dots, H_{r-1}/N$, where $H_i$ is a normal subgroup of $G$ containing $N$, and where $|H_i/N|=p^i$.
So, $|H_i|/|N|=p^i$ and thus $|H_i|=p^i|N|=p^ip=p^{i+1}$. So, there exists a normal subgroup of order $$|\{1\}|=p^0, |H_0|=p^1, |H_1|=p^2, \dots, |H_{r-1}|=p^r.$$