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I have been given the following problem and starting steps but I am stuck on the first step

A measure of goodness of fit to estimate the mean $\mu$ and standard deviation $\sigma$ of a normally distributed random variable is maximum likelihood estimation. For sample $y = (y_1,... y_n)$ drawn from the normally distributed random variables, $Y_i \sim > N(\mu,\sigma)$, the likelihood is given by
$p(y|\mu,\sigma)=\prod_{i=1}^{n}\big(\frac{1}{2\pi\sigma^2}\big)^{\frac{1}{2}} exp \big( - \frac{1}{2\sigma^2}(\mu-y_i)^2 \big) $
Using the fact that $e^{-a}e^{-b} = e^{-a-b}$ for arbitrary variables a and b, show the negative log likelihood has the following form:
$L_(y|\mu,\sigma) = -\log p(y|\mu,\sigma) +\frac{1}{2\sigma^2}\sum_{i=1}^n(y_i-\mu)^2$

The steps start with splitting up the products to be
$\prod_{i=1}^n(\frac{1}{2\pi\sigma^2})^{\frac{1}{2}}\prod_{i=1}^n exp\big( - \frac{1}{2\sigma^2}(\mu-y_i)^2 \big)$

I don't see why this can be done because

$(a1 *a2) (b1* b2) \ne (a1* a2) b1 ( a1 *a2) b2$

What am I missing?

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Someone explained that I could think of it this way:
Consider the sum operator
$\sum_{i=1}^n a+y_i = \sum_{i=1}^na+\sum_{i=1}^n y_i $

The product operator can work similarly
$\prod_{i=1}^n a y_i = \prod_{i=1}^na\prod_{i=1}^n y_i $

I guess I am just not used to working with the product operator.

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