smc

If we change the base of $Lo{g}_{M}N$ to base $N$, we get $\frac{1}{lo{g}_{N}M}=Lo{g}_{N}M$. Multiplying both sides by $Lo{g}_{N}M$, we get $Lo{g}_N{M}^2=1$. Since $N\ne M$, $Lo{g}_{N}M=-1$. So $N^{-1}=M\to \frac{1}{N}=M\to MN=1$. So the answer is $\boxed{1}$.