A question on an identity involving partition Let $n$ be a natural number. Let $\lambda \mapsto n$ , be a partition of n. So $\lambda=(\lambda_1, \lambda_2, \ldots ,\lambda_k)$, with $\lambda_1\leq \ldots \leq \lambda_k$, and $\sum_{i=1}^k \lambda_i=n$. Now, let $m_i(\lambda)$ denote the number of times $i$ occur in the partition $\lambda$. Consider the following 
$$\sum_{\lambda \mapsto n} \frac{1}{\prod_{i=1}^{n} i^{m_i(\lambda)}m_i(\lambda) !}=1$$.
I want to know whether this identity is correct. If it is true, what can be a proof of this equality!
Thank you.
 A: 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.
