Let $\zeta$ be a primitive $p^2$ th root of unity with prime $p$. Let $L$ be the Galois extension $\mathbb{Q}(\zeta)$,
I want to prove $L$ contains a subfield whose Galois group is $\mathbb{Z}/p\mathbb{Z}$.
The followings are my trials stuck at some point.
First I know, $\mathbb{Q}(\zeta_n)$ where $\zeta_n$ is the $n$th root of unity, is Galois over $\mathbb{Q}$ and $\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \simeq (\mathbb{Z}/n \mathbb{Z})^{*}$, where $*$ denote the unit group.
Hence the $\mathbb{Q}$ automorphism of $L$ is the form $\zeta \mapsto \zeta^k$ with $p$ does not divide $k$.
To show $L$ contains a subfield whose Galois group is $\mathbb{Z}/p\mathbb{Z}$, first let $K$ be a subfield of $L$, then having $\operatorname{Gal}(L/K) \simeq \mathbb{Z}/p\mathbb{Z}$ implies $[L:K]=p$, .....
Is there any simple way to prove this theorem?
Simple depends.
It is extremely simple with some standard factors about the unit groups $(\mathbb Z/n\mathbb Z)^\ast$. The relevant here being:
$$ (\mathbb Z/p^n\mathbb Z)^\ast\cong\begin{cases}(\mathbb Z/p\mathbb Z)^\ast\times\mathbb Z/p^{n-1}\mathbb Z&,p>2\\\mathbb Z/2\mathbb Z\times\mathbb Z/2^{n-2}\mathbb Z&,p=2\end{cases} $$
In our case, $n=2$, we the immediately deduce that
$$ \operatorname{Gal}(\mathbb Q(\zeta)/\mathbb Q)\cong\begin{cases}(\mathbb Z/p\mathbb Z)^\ast\times\mathbb Z/p\mathbb Z&,p>2\\\mathbb Z/2\mathbb Z&,p=2\end{cases} $$
which solves the exercise.
Alternatively, note that $\operatorname{Gal}(\mathbb Q(\zeta)/\mathbb Q)$ is abelian. Hence the structure theorem applies, giving that
$$ \operatorname{Gal}(\mathbb Q(\zeta)/\mathbb Q)\cong\mathbb Z/p\mathbb Z\oplus\left(\bigoplus_i\mathbb Z/q_i^{e_i}\mathbb Q\right) $$
with $q\mid p-1$. The decomposition takes this form since $(p,p-1)=1$. Fixing by the right-most summand gives the result.
Both of the given solutions, however, require some outside knowledge (i.e. beyond the scope of the question itself). There might be an easier, group-theoretic argument which I do not see at the moment.
EDIT: I found an interesting alternative (which only works for $n=2$ and $p>2$; $p=2$ is not that hard though). Consider taking $p$-th powers as an endomorphism $\pi$ of $(\mathbb Z/p^2\mathbb Z)^\ast$. Note that
$$ \pi^2(a)\equiv a^{p^2}\equiv a^p\equiv\pi(a)\mod p^2 $$
by Euler's theorem. Hence $\pi^2=\pi$ and it is then straightforward to see that
$$ (\mathbb Z/p^2\mathbb Z)^\ast\cong\ker\pi\oplus\operatorname{im}\pi\,. $$
Note that $\ker\pi$ consists of elements of order $p$ and that $\ker\pi\ne1$ since $p+1$ is non-trivial of order $p$ (alternatively, using Cauchy's theorem, since $p$ divides the group order). As $|(\mathbb Z/p^2\mathbb Z)^\ast|=p(p-1)$ we conclude that $\ker\pi\cong\mathbb Z/p\mathbb Z$ and fixing by $\operatorname{im}\pi$ yields the desired conclusion.