Notation:
(i) $\uparrow^{G}_{H}$ denotes the induced representation (or its corresponding character) of $H$ to $G$, and $\downarrow^{G}_{H}$ denotes restriction similarly.
(ii) $\lambda = \displaystyle\prod i^{m^{(\lambda)}_{i}}$ denotes that $\lambda$ is a partition where $i$ is repeated $m^{(\lambda)}_i$ times, i.e. there exists exactly $m^{(\lambda)}_i$ $k$s for which $\lambda_k = i$.
(iii) $\mathfrak{S}_{n}$ is the symmetric group on $n$ letters, and $\mathfrak{S}_\lambda$ is the Young subgroup associated to the partition $\lambda$.
(iv) $K_{\lambda\mu}$ is the Kostka number associated to the partitions $\lambda$ and $\mu$.
(v) $C_{\nu}$ is the conjugacy class of $\mathfrak{S}_n$ consisting of elements of cycle type $\nu$.
(vi) $z_{\nu} = n!/|C_\nu| = \displaystyle\prod_{i}i^{m^{(\nu)}_i}m^{(\nu)}_i!$ is the size of the centralizer of an element in $\mathfrak{S}_n$ of cycle type $\nu$.
Theorem. $\displaystyle\prod_{\lambda\vdash n} \displaystyle\prod_{k}\lambda_{k} = \displaystyle\prod_{\lambda\vdash n} \displaystyle\prod_{i} m^{(\lambda)}_i!$.
I encountered the above identity while studying representation theory of symmetric groups. I give my proof below:
Fix $n$ and order the partitions of $n$ by lexicographic order.
Given $\lambda \vdash n$, let $\phi_\lambda$ be the character of the permutation representation $1{\uparrow}^{\mathfrak{S}_{n}}_{\mathfrak{S}_{\lambda}}$, and $\chi_\lambda$ the character of the Specht module $S^{\lambda}$.
Define matrices $\mathcal{A}_{\lambda\mu} = z^{1/2}_\mu|\mathfrak{S}_\lambda \cap C_\mu|$ and $\mathcal{B}_{\lambda\mu} = |\mathfrak{S}_\mu|\cdot K_{\lambda\mu}$, then we have:
$\begin{align*} (\mathcal{B}^{t}\mathcal{B})_{\lambda\mu} &= \displaystyle\sum_{\nu}|\mathfrak{S}_\lambda|\cdot K_{\nu\lambda}\cdot |\mathfrak{S}_\mu|\cdot K_{\nu\mu} \\ &= |\mathfrak{S}_\lambda||\mathfrak{S}_\mu| \displaystyle\sum_{\nu}\langle\phi_\lambda,\chi_\nu\rangle \langle\chi_\nu,\phi_\mu\rangle \quad\text{(Young's Rule)}\\ &= |\mathfrak{S}_\lambda||\mathfrak{S}_\mu| \langle\phi_\lambda,\phi_\mu\rangle \quad{(\text{orthonormality of }\chi_\nu)}\\ &= |\mathfrak{S}_\lambda||\mathfrak{S}_\mu| \langle1{\uparrow}^{\mathfrak{S}_n}_{\mathfrak{S}_\lambda}{\downarrow}^{\mathfrak{S}_n}_{\mathfrak{S}_\mu},1_{\mathfrak{S}_\mu}\rangle \quad\text{(Frobenius Reciprocity)}\\ &= |\mathfrak{S}_\lambda|\displaystyle\sum_{\nu}\phi_\lambda(C_\nu) \cdot |\mathfrak{S}_\mu \cap C_\nu| \\ &= \displaystyle\sum_{\nu}z_\nu|\mathfrak{S}_\lambda \cap C_\nu||\mathfrak{S}_\mu \cap C_\nu| \quad\text{(character of permutation module)} \\ &= (\mathcal{A}\mathcal{A}^{t})_{\lambda\mu} \end{align*}$
Observe that $\mathcal{A}$ and $\mathcal{B}$ are upper triangular (straightforward for $\mathcal{A}$, and for $\mathcal{B}$ since $K_{\lambda\mu} = 0$ unless $\lambda$ dominates $\mu$), thus the determinants of both matrices are just the products along the diagonal:
$\det \mathcal{A} = \displaystyle\prod_{\lambda\vdash n} z^{1/2}_\lambda |\mathfrak{S}_\lambda \cap C_\lambda| = \displaystyle\prod_{\lambda\vdash n}z^{1/2}_\lambda\displaystyle\prod_{k}(\lambda_k-1)!$
$\det \mathcal{B} = \displaystyle\prod_{\lambda\vdash n} |\mathfrak{S}_\lambda| \cdot K_{\lambda\lambda} = \displaystyle\prod_{\lambda\vdash n}\displaystyle\prod_{k}\lambda_k!$
Finally, expanding $(\det \mathcal{A})^2 = (\det \mathcal{B})^2$ yields:
$\displaystyle\prod_{\lambda\vdash n}z_\lambda\displaystyle\prod_{k}(\lambda_k-1)!^2 = \displaystyle\prod_{\lambda\vdash n}\displaystyle\prod_{k}\lambda_k!^2$
$\displaystyle\prod_{\lambda\vdash n}z_\lambda = \displaystyle\prod_{\lambda\vdash n}\displaystyle\prod_{i}i^{m^{(\nu)}_i}m^{(\nu)}_i! = \displaystyle\prod_{\lambda\vdash n}\displaystyle\prod_{k}\lambda_k^2$
$\displaystyle\prod_{\lambda\vdash n}\displaystyle\prod_{i}m^{(\nu)}_i! = \displaystyle\prod_{\lambda\vdash n}\displaystyle\prod_{k}\lambda_k$
The identity itself is simple enough that I suspect there may be a purely combinatorial proof. In fact:
For $n = 4, 5, 6, 7$, the multisets $\{\lambda_k \mid \lambda\vdash n \}$ and $\displaystyle\bigcup_{\lambda\vdash n}\displaystyle\bigcup_i\, [m^{(\lambda)}_i]$ are identical (where $[m]$ denotes $\{1,2, ..., m\}$),
i.e. when you expand the factorials $m!$ into a product of numbers from $1$ to $m$, there is a bijection between the numbers occurring in each product.
For $n=6$, we have $\lambda = 6, 51, 42, 411, 33, 321, 3111, 222, 2211, 21111, 111111$.
Abusing notation, write $[m^{(\lambda)}] = 1, 11, 11, 121, 21, 111, 1321, 321, 2121, 14321, 654321$, respectively for each $\lambda$ above. (For example, $[m^{3111}] = 1321$ since there is $1$ $3$ and $3$ $1$s; $[1] = \{1\}, [3] = \{3,2,1\}$.)
Now, counting the numbers occurring in the lists above for $\lambda$ and $[m^{(\lambda)}]$, we have $19$ $1$s, $8$ $2$s, $4$ $3$s, $2$ $4$s, $1$ $5$ and $1$ $6$ in each.
Does the above bijection hold for all $n$? If so, is there a direct bijective proof using the above idea?
If not, is there at least a representation theory free proof?
Note: A question asking for a proof of this identity is already here (which, I guess, I technically have provided an answer to), but I wasn't able to find the corresponding problem on EC2, nor am I able to provide a combinatorial interpretation, which is why I ask it again with more details provided.
Edit: Here is a completion of the outline given in JBL's answer into the desired bijection of multisets. Fix a positive integer $j$ and consider the following sequence of bijections:
$\begin{align*} \{(\lambda,i)\vdash n \mid \lambda_i \text{ is the } j\text{th } k\} &\longleftrightarrow \{\lambda\vdash n \mid m^{(\lambda)}_k \geq j\} \qquad\text{(}i\text{ is unique for fixed }j)\\ &\longleftrightarrow \{\lambda_-\vdash (n-jk) \} \qquad\text{(delete }j\text{ parts of size }k)\\ &\longleftrightarrow \{\lambda'\vdash n \mid m^{(\lambda')}_j \geq k\} \qquad\text{(insert }k\text{ parts of size }j) \end{align*}$
Observing that marked partitions correspond to parts, we obtain a bijection between parts $\lambda_i$ that are $j$th $k$s of some partition $\lambda$ and $m^{(\lambda')}_j$s that are at least $k$ for another partition $\lambda'$. Union over $j$ gives us the full bijection between multisets.
I am quite confident that this was assigned to me as a graduate student in one of Richard's classes; therefore I expect that it is in EC1, chapter 1 exercises somewhere. Nevertheless, here is a quick outline of a combinatorial proof: say that a pair $(\lambda, i)$ is a $k$-marked partition if $\lambda_i = k$. It is easy to see that the number of parts of size $k$ among all partitions of $n$ is equal to the number of $k$-marked partitions of $n$. Moreover, it is not hard to see that the number of $k$-marked partitions in which $i$ is the $j$th part of size $k$ is exactly $p(n - jk)$: there's a bijection that amounts to adding or removing a particular $j \times k$ rectangle in the Young diagram.
Now let's consider the other (multiplicities) side. Given a partition $\lambda$, it contributes a copy of $k$ for each part size that appears with multiplicity at least $k$. Well, the number of partitions of $n$ in which $j$ appears with multiplicity at least $k$ is exactly $p(n - jk)$: a bijection that amounts to adding or removing a particular $k \times j$ rectangle in the Young diagram.