Many Dices!

For the product of the numbers shown by $N$ dice to be even, we just need one die to show an even number. The only way this doesn't happen is if all the dice show odd numbers. Each die shows an odd number with probability $\frac12$ ; so the probability of them all being odd is $\frac{1}{2^N}$

And the probability the product is even is $1-\frac{1}{2^N}=\frac{2^N-1}{2^N}$

This fraction is already in lowest terms, so the answer is $2^N- \left(2^N-1 \right)=\boxed1$ .