Let $\lambda$ be a partition of $N$ of rank $r$. How can I show that:
$$\sum_wx^\lambda(w)=f^\lambda(-1)^{t(\lambda)}\prod^r_{i=1}(\lambda_i-1)!(\lambda'_i-1)!$$
where $w$ ranges over all permutations in $\mathfrak{S}_N$ with exactly $r$ cycles, and where $t(\lambda)=\sum^r_{i=1}(\lambda'-1)$.
I'm not trying to ask for someone to do this for me, I'm just asking for a general hint.
Thanks for the help!
For any $\lambda\in Par$ and $n\in\mathbb{P}$:
$$s_\lambda(1^n)=\prod_{u\in\lambda}\frac{n+c(u)}{h(u)}$$
where we define the hook length $h(u)$ of $\lambda$ at $u$ by $h(u)=\lambda_i+\lambda'_j-i-j+1$ and the content $c(u)$ of $\lambda$ at $u=(i,j)$ by $c(u)=j-i$.
We know that:
$$\frac{1}{N!}\sum_{w\in\mathfrak{S}_N}x^\lambda(w)n^{c(w)}=\prod_{u\in\lambda}\frac{n+c(u)}{h(u)}$$
The coefficient of $n^r$on LHS is equal to $\frac{1}{N!}\sum_wx^\lambda(w)$. Thus we have $x^\lambda(v)=0$. So the LHS is divisible by $n^r$. There're $r$ factors equal to $n$ in the numerator of LHS, coming from that the $r$ diagonal terms $u=(i,i)$.
So, the coefficient of $n^r$ on RHS would be:
$$\frac{\prod_{u\in\lambda,u\neq(i,i)}c(u)}{\prod_{u\in\lambda}h(u)}=\frac{f^\lambda}{N!}\prod_{u\in\lambda, u\neq(i,i)}c(u).$$
Now we can see that this last product is simply $$(-1)^{t(\lambda)}\prod^r_{i=1}(\lambda_i-1)!(\lambda'_i-1)!.\ \ \ \ \square$$