Why does $\mu$ and $\sigma$ minimise the product of normal distributions? First of all, there might be a better way to phrase the question, this is in a course about real analysis so I'm not sure if there's a better name for this function that is clearly related to probability somehow.
Let $x_1,\ldots,x_N$ be fixed real numbers. I am asked to find values of $\mu$ and $\sigma$ which minimize the function:
$$\begin{pmatrix} \mu \\ \sigma \end{pmatrix} \mapsto \prod_{i=1}^N \frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{1}{2\sigma^2}(x_i - \mu)^2\right).$$
Obviously the expression in the the product is the probability density function of a normal distribution with mean $\mu$ and standard deviation $\sigma$, so I expect the punchline to be:
$$\mu = \frac{\sum x_i}{N}, \qquad \sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{N}}.$$
Simplifying the product I get:
$$ \left(\frac{1}{\sqrt{2\pi}\sigma}\right)^N \exp\left( -\frac{1}{2\sigma^2} \sum_{i=1}^N(x_i - \mu)^2\right),$$
from which I can accept that choosing $\mu$ to be the mean is the "best" way to minimise this expression (although rigorously proving this would be beyond me).
If I now assume that $\mu$ is the mean and $\sigma$ is the standard deviation, my expression becomes:
$$ \left(\frac{1}{\sqrt{2\pi}\sigma}\right)^N \exp\left(-\frac{1}{2}\frac{N\sum (x_i - \mu)^2}{\sum (x_i - \mu)^2}\right),$$
or:
$$ \left(\frac{1}{\sqrt{\frac{2\pi}{N}\sum (x_i - \mu)^2}}\right)^N\exp\left(-\frac{N}{2}\right).$$
I have no idea how to prove that this is minimal. It certainly makes the exponent a lot simpler to look at, but beyond that I am unsure how to answer.

The expressions you've got for $\mu$ and $\sigma^2$ are maximum points and not minimum points. The traditional way to solve this is applying log to the original expression and to obtain a concave function. Therefore a solution for the first order condition (null gradient) is a global max. And the expressions for $\mu$ and $\sigma^2$ are exactly this solution. They are called the "maximum likelihood" estimators in this context.
