TITLE: Murnaghan-Nakayama rule when all cycles have same size
QUESTION [5 upvotes]: Let $\lambda \vdash nk$. Let $n^k$ denote the partiton with $k$ parts of size $n$. We can compute $\chi^\lambda(n^k)$ by using the Murnaghan-Nakayama rule, as a signed sum over border-strip tableaux, (shape $\lambda$ and border strips of size $n$). 
Now, by checking several cases on the computer, it seems like each term in the (Murnaghan-Nakayama) sum has the same sign. Combinatorially, it reduces to proving the following:
Define the height of a border-strip to be the maximal row index that it touches, minus the minimal row index. The height of a border-strip decomposition of the diagram $\lambda$  is the sum of the heights of the border-strips.  Show that the parity of this sum is independent of the actual decomposition (it only depends on $\lambda$ and $n$).
I am looking for a reference of this (I do think I can in fact prove a stronger statement, but I have not worked out the details yet).

REPLY [3 votes]: This is corollary 10 in "A bijection proving orthogonality of the characters of $S_n$", by Dennis E. White.