Let $G$ be a group of order $p^n$. I have to show that there are normal subgroups of $G$ of orders $1,p,...,p^{n-1},p^n$.
My attempt :
The trivial subgroup is a normal subgroup of order $1$. Assume $G$ has a normal subgroup $N$ of order $p^k,k<n$. To show: $G$ has a normal subgroup of order $p^{k+1}$.
Let $\gamma : G \rightarrow G/N$ be the natural homomorphism. The kernel is $N$, whose size is $p^k$. So $\gamma$ is $p^k-\text{to}-1$. Since $k<n$, $G/N$ is also a $p$-group. Hence, $G/N$ has a non-trivial centre. $Z(G/N)$ is then also a $p$-group. So by Cauchy's theorem, it has an element $a$ of order $p$. The subgroup $\langle a\rangle$ generated by this element is a subgroup of $Z(G/N)$. Therefore, it is normal in $G/N$. Then the pullback of $\langle a\rangle$ is normal in $G$. Since the order of $\langle a\rangle$ is $p$, the order of $\gamma^{-1}\langle a\rangle$ is $p^k×p=p^{k+1}$. Thus we have found a normal subgroup of $G$ of order $p^{k+1}$. This completes my proof.
Your answer is partially correct: you have made use of non-triviality of center in $p$-groups after passing to $G/N$, but not for $G$. In other words, you took $N$ a normal subgroup of order $p^k$ and $G\rightarrow G/N$ the natural homomorphism; but how do you know that $N\neq 1$? IF $N=1$ then the homomorphism is no longer helpful to carry out induction.
In short, the existence of normal subgroups of every order follow from following two facts:
For a $p$-group $G\neq 1$, the center is non-trivial.
If $N\trianglelefteq G$ and $G\rightarrow G/N$ is natural homomorphism then there is a (nice) bijection between normal subgroups of $G/N$ and the normal subgroups of $G$ which contain $N$.