jmc

Since $N!=(N-1)!(N)$, we can rewrite the given fraction as $\frac{N!}{(N+1)!}$. We can rewrite the denominator as $(N+1)!=(N!)(N+1)$, so the fraction becomes $\frac{N!}{(N!)(N+1)}$. Canceling an $N!$ from the numerator and denominator, we are left with $\boxed{\frac{1}{N+1}}$.