For some reason I am having an extremely hard time finding out how the following expression is derived $$ \hat{\beta}_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} $$
Is there some kind of identity that I am missing in order to derive this expression?
I've tried plugging in $\hat{\beta}_0 = \bar{y} - \hat{\beta}_1 \bar{x}$ in to $\sum x_i y_i = \hat{\beta}_0 \sum x_i + \hat{\beta}_1 \sum x_i^2$, which gives me $$ \hat{\beta}_1 = \frac{\sum x_iy_i - \bar{y} \sum x_i}{\sum x_i^2 - \bar{x} \sum x_i} $$ but I do not see how I should proceed from here. I feel that there should be an identity linking $\bar{x}$ or $\bar{y}$ to $\sum x_i y_i$ or $\sum x_i^2$, but nothing comes to mind at the moment.
You may observe that $$ \hat{\beta}_1 =\frac{\sum_{i=1}^n x_i y_i - n \bar{x}\bar{y}}{\sum_{i=1}^n x_i^2 -n\bar{x}^2} =\frac{\frac1n\sum_{i=1}^n x_i y_i - \bar{x}\bar{y}}{\frac1n\sum_{i=1}^n x_i^2 -\bar{x}^2}.\tag1 $$ Then you may prove that $$ \frac1n\sum_{i=1}^n \left(x_i -\bar{x}\right)^2=\frac1n\sum_{i=1}^n x_i^2 -\bar{x}^2 \tag2 $$ and that $$ \frac1n\sum_{i=1}^n \left(x_i -\bar{x}\right)\left(y_i -\bar{y}\right)=\frac1n\sum_{i=1}^n x_iy_i -\bar{x}\bar{y}. \tag3 $$
Let's see a proof of $(2)$. We have $$ \begin{align} \frac1n\sum_{i=1}^n \left(x_i -\bar{x}\right)^2&=\frac1n\sum_{i=1}^n \left(x^2_i -2x_i\bar{x}+\bar{x}^2\right) \\\\&=\frac1n\sum_{i=1}^n x^2_i -2\bar{x}\frac1n\sum_{i=1}^nx_i+\frac1n\sum_{i=1}^n\bar{x}^2 \\\\&=\frac1n\sum_{i=1}^n x^2_i -2\:\bar{x}\times \bar{x}+\frac1n \times n\:\bar{x}^2 \\\\&=\frac1n\sum_{i=1}^n x^2_i -\bar{x}^2. \end{align} $$ Now use a similar path to get $(3)$.
Thus, you obtain $$ \hat{\beta}_1 =\frac{\sum_{i=1}^n x_i y_i - n \bar{x}\bar{y}}{\sum_{i=1}^n x_i^2 -n\bar{x}^2} =\frac{\frac1n\sum_{i=1}^n x_i y_i - \bar{x}\bar{y}}{\frac1n\sum_{i=1}^n x_i^2 -\bar{x}^2}=\frac{\frac1n\sum_{i=1}^n \left(x_i -\bar{x}\right)\left(y_i -\bar{y}\right)}{\frac1n\sum_{i=1}^n \left(x_i -\bar{x}\right)^2}.\tag4 $$