I'm having difficulty understanding this proof: $$ \begin{align} E[\bar{X}] &= E[\frac{X_{1} + ... + X_{n}}{n}] \\ &= \frac{1}{n}(E[X_{1}] +...+E[X_{n}]) \\ &= \mu \end{align}$$
Wouldn't the second line represent $$ \frac{1}{n}(E[X_{1}] +...+E[X_{n}]) = \frac{1}{n}(\frac{X_{1} + ... + X_{n}}{n}) =\frac{X_{1} + ... + X_{n}}{n^2}$$ And if so, I'm confused as to how this represents the population mean.
I am assuming that $E[X] = \mu$, so then by linearity of expectation, $$E[X_1+...+X_n] = E[X_1 ]+ ...+ E[X_n] =n\mu$$ so obviously $\frac{1}{n} E[X_1+...+X_n]$ will give just $\mu$.