Consider the next identity $$ \sum_{\tau \vdash r} d^2 (\tau ) \! \prod_{i = -(r-1)}^{r-1} \! \! \left( N+i \right)^{t_i^r} =\sum_{\tau \vdash r} d^2 (\tau ) \prod_{i = 1}^{r} \Gamma [N + \tau_i - i +1]= r! \, N^{r} $$ where $d(\tau)$ is dimension of representation of a permutation group $S_r$ and $\tau =(\tau_1,...,\tau_{l(\tau)})$ is a partition of $\tau \vdash r$, some power`s coefficients $t_i^r$ dictated by corresponding Young diagram.
I don`t have proof or combinatorial meaning of this fact.
For example $r=2$ $$ N(N+1)+(N-1)N= 2! N^2 \ , $$ for $r=3$ $$ N(N+1)(N+2)+2^2(N-1)N(N+1)+(N-2)(N-1)N= 3! N^3 \ , $$ for $r=4$ we have less trivial case $$ N(N+1)(N+2)(N+3)+3^2(N-1)N(N+1)(N+2)+ 2^2(N-1)N^2(N+1)+3^2(N-2)(N-1)N(N+1)+(N-3)(N-2)(N-1)N = 4!\, N^4 $$ and so on.
The structure of the summand in brackets is cleared by the illustration of the Young diagram for partition $(3,2,2)$ that belongs to $r=7$ case.
Even don`t clear, is this identity obvious. Is this a known fact at all?
P.S. Of course, this is more sophisticated modification of known identity $$ \sum_{\tau \vdash r} d^2 (\tau ) =r! $$
