Well, I have an answer with the explanation:

consider a N-ary tree with depth d

Total no of nodes in the tree, T=1+N+N^2+...+N^d= (1-N^(d+1))/(1-N) --(1)

No. of non leaf nodes, M=(1-N^d)/(1-N) ---(2)

No. of leaf nodes = T-M = N^d ---(3)

from eq (2), d= log(base N) [MN-M+1] ---(4)

Therefore, no of leaf nodes = N^d = MN-M+1