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?