Observe that the cycle index of the symmetric group is given by
$$Z(S_n) = \sum_{\lambda \vdash n}
\frac{1}{\prod_{i=1}^n i^{m_i(\lambda)} m_i(\lambda)!}
\prod_{i=1}^n a_i^{m_i(\lambda)}.$$
This is the unlabeled multiset combinatorial class
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small#2}}}\textsc{MSET}_{=n}.$$
Now we may use Burnside or PET. With the former when setting all $a_i$
to one i.e. computing
$$Z(S_n; 1,1,1,\ldots)$$
we obtain the number of colorings of $n$ interchangeable slots with
one color. There is just one such coloring, proving the claim.
Alternatively using PET we count the number of multisets of $n$
elements drawn from a source of one element, which is
$\textsc{MSET}_{=n}(\mathcal{Z})$ with generating function
$$\sum_{\lambda \vdash n}
\frac{1}{\prod_{i=1}^n i^{m_i(\lambda)} m_i(\lambda)!}
\prod_{i=1}^n z^{i\times m_i(\lambda)}
= z^n \sum_{\lambda \vdash n}
\frac{1}{\prod_{i=1}^n i^{m_i(\lambda)} m_i(\lambda)!}.$$
There is just one such multiset and we must have $f(z) = z^n$, proving
that the scalar is one, and we once more have the claim.
The underlying principle here is that
$$\frac{n!}{\prod_{i=1}^n i^{m_i(\lambda)} m_i(\lambda)!}
= \frac{n!}{\prod_{i=1}^n (i!)^{m_i(\lambda)}}
\prod_{i=1}^n \left(\frac{i!}{i}\right)^{m_i(\lambda)}
\prod_{i=1}^n \frac{1}{m_i(\lambda)!}$$
counts the number of permutations in the symmetric group on $n$
elements with the lengths of the cycles in the factorization of such a
permutation into disjoint cycles being given by the elements of
$\lambda$ i.e.
$\prod_{i=1}^n a_i^{m_i(\lambda)}.$
With this classification we obtain every permutation exactly once and
the sum of the LHS over $\lambda\vdash n$ is indeed equal to $n!$ as
remarked in the comments.
