An identity related to partitions into $n$ parts and Schur polynomials While working with Schur polynomials I found what seems like a nice identity, and I wonder if it has a simple proof.
Notation: Suppose $d,n\in\mathbb{N}$, and $\lambda =(\lambda_1,\dots,\lambda_n)$ is an ordered partition of $d$, that is
$\lambda_1 \ge \dots \ge \lambda_n \ge 0$, and $\lambda_1 + \dots + \lambda_n = d$.
Let $\Lambda(d,n)$ be the set of all such partitions. For each $\lambda \in \Lambda(d,n)$ define:
$$ N(\lambda) = \prod_{1\le i < j \le n} \frac{\lambda_i - \lambda_j + j - i}{j - i},
$$
and
$$ W(\lambda) = \prod_{1 \le i \le n} (\lambda_i + n - i)!.
$$
Remark: $N(\lambda)=s_\lambda(1,\dots,1)$, where $s_\lambda$ is the Schur polynomial associated with the partition $\lambda$.
Now, define
$$ A(d,n) = \sum_{\lambda \in \Lambda(d,n)} \frac{N(\lambda)^2}{W(\lambda)}.
$$
It seems the following identity holds:
$$ A(d,n) = \left( \prod_{k=0}^{n-1} k! \right)^{-1} \frac{n^d}{d!}.$$
Is there a simple proof\explanation?
 A: I found a proof which I don't really like, but I'll share it.
For two (real) diagonal matrices $A,B$, the Harish-Chandra-Itzykson-Zuber (HCIZ) integral is
$$
I(A,B) = \int_{U(n)} e^{\rm{tr}(U^* A U B)} \, \rm{d} U = c_n \frac{\det\left([e^{a_j b_k}]_{j,k=1}^n\right)}{\Delta(a)\Delta(b)},
$$
where $\Delta(a) = \prod_{j<k} (a_k - a_k)$ is the Vandermonde determinant, and $c_n = \Delta([1,\dots,n]) = \prod_{k=1}^{n-1} k!$ (the integration is with respect to Haar measure on the unitary group).
Using an infinite version of the Cauchy-Binet formula, we can write
$$
\det\left(\{e^{a_j b_k}\}_{j,k=1}^n\right) = \sum_\lambda \det\left(\left[\frac{a_j^{\lambda_k+n-k}}{\sqrt{(\lambda_k+n-k)!}}\right]_{j,k=1}^n\right)
\det\left(\left[\frac{b_k^{\lambda_j+n-j}}{\sqrt{(\lambda_j+n-j)!}}\right]_{j,k=1}^n\right),
$$
where the sum is over all partitions $\lambda$ of size $n$. Using the fact
$$ s_\lambda(a) = \frac{\det\left(a_j^{\lambda_k+n-k}\right)}{\Delta(a)},$$ we find
$$ I(A,B) = c_n \sum_\lambda \frac{s_\lambda(a)s_\lambda(b)}{\prod_{j=1}^n (\lambda_j+n-j)!}
= c_n \sum_\lambda \frac{s_\lambda(a)s_\lambda(b)}{W(\lambda)}.
$$
If we substitute $A = B = t \cdot \rm{Id}_n$, then we find
$$
e^{t^2 n} = c_n \sum_{d\ge 0} t^{2d} \sum_{\lambda\in\Lambda(d,n)} \frac{\left(s_\lambda(1,\dots,1)\right)^2}{W(\lambda)}.
$$
We get the identity by comparing coefficients of $t$.
A: We can use the fact that $N(\lambda)=\left|\text{SSYT}(\lambda)\right|$, the number of semistandard Young tableaux of shape $\lambda$, and that $d!\cdot\left(\frac{\prod_{1\le i < j\le n}(\lambda_i - \lambda_j + j - i)}{\prod_{1 \le i \le n} (\lambda_i + n - i)!}\right)=\left|\text{SYT}(\lambda)\right|$, the number of standard Young tableaux of shape $\lambda$. By some simple rearranging your identity becomes
$$\sum_{|\lambda|=d}\left|\text{SSYT}(\lambda)\right|\left|\text{SYT}(\lambda)\right|=n^d.$$
Now, if you track the left hand side through the Robinson-Schensted-Knuth correspondence, you'll find that it corresponds to $d\times n$ matrices with row sums all $1$. So there are exactly $n^d$ of these.
