Converting product multiplier of exponential to summation

Question

Why is

$$\prod_{n=1}^N (\exp(x_n - y)) = \exp\left( \sum_{n=1}^N (x_n)-Ny\right) $$

How does the product operator with exponential break out into a summation operation with the exp outside?

Answer 1

$$\prod_{n=1}^N \left(\exp(x_n-y)\right)=(\exp(x_1-y)) (\exp(x_2-y))\cdots (\exp(x_N-y))$$ But $e^{a_1}e^{a_2} \cdots e^{a_n}= e^{a_1 + a_2 + \cdots + a_n}$ SO $$\prod_{n=1}^N (\exp(x_n-y)) = \exp\left( \sum\limits_{i=1}^N (x_i - y)\right) \tag{1}$$ But $$\sum\limits_{i=1}^N (x_i - y)= \sum\limits_{i=1}^N x_i -\sum\limits_{i=1}^N y$$ But $y$ is a constant so $$\sum\limits_{i=1}^N (x_i - y)= \sum\limits_{i=1}^N x_i -Ny$$ Replace in equation $(1)$ and the proof is done

Answer 2

Remember when you multiply exponential, exponents add up.$$ e^{ x_1-y} e^{x_1-y} \cdots e^{ x_N-y} = e^{x_1+x_2+\cdots+x_N -Ny}$$