TITLE: $N^L$ vs. ${N+L\choose N}$
QUESTION [5 upvotes]: Any ideas on finding a good estimate/approximation for $\frac AB$ where $A = N^L$ and $B = {N+L\choose N}$?

REPLY [3 votes]: $\log(A/B) = \log(L!) - \sum_{j=1}^L \log(1+j/N)$.  You can approximate or bound the sum in various ways, depending on your needs.