How to show that sum of squares is n times of mean?

I gave $X_i\sim N(\mu=8, \sigma^2=1)$ for $i=1,...91$ with observed $\bar{x}=7.319$ and I calculate $f(\bar{x}=7.319|\mu_0=8)$ and I stuck in one step of calculation, actually the very first:

$(\frac{1}{\sigma\sqrt{2\pi}})^n \exp(-\frac{\sum (x_i-\mu_0)^2}{2\sigma^2})= (\frac{1}{\sigma\sqrt{2\pi}})^n \exp(-\frac{n (\bar{x}-\mu_0)^2}{2\sigma^2})$

How come that: $\sum(x_i - \mu_0)^2 =n(\bar{x}-\mu_0)^2$?

Please, help, I've been thinking and calculating for hours and still I can't reach the right side from the left.

The equality holds if and only if $x_1 = x_2 = \cdots = x_n$, by Cauchy-Schwarz inequality: $(\sum(x_i-\mu))^2\leq n\sum(x_i-\mu)^2$.