Was reading Wikipedia's proof that the maximum likelihood estimator of variance is biased. Here is a link.
I follow the proof until the very last step. We have, $$\sigma^2-E[(\bar{X}-\mu)^2].$$ Noting that $\bar{X}$ is sample mean we rewrite this as, $$\sigma^2-E\bigg[\bigg(\frac{1}{n}\sum_{i=1}^{n}X_i+\frac{1}{n}\mu\bigg)^2\bigg],$$ but I don't see how it follows that this is equivalent to $\frac{1}{n}\sigma^2$.
Recall that for some r.v. $X$ $$ \operatorname{var}(X) = \mathbb{E}( X - \mathbb{E}[X] ) ^ 2, $$ and $\mathbb{E}\bar{X}_n = \mu$, hence $$ \mathbb{E}( \bar{X}_n - \mu ) ^ 2 = \operatorname{var}(\bar{X}) = \frac{\sigma^2}{n} . $$