I know there are some proof in the internet, but I attempted to proove the formulas for the intercept and the slope in simple linear regression using Least squares, some algebra, and partial derivatives (although I might want to do it wituout partials if it's easier).
I've posted my attempt below. I don't know what to from here.
The principle underlying least squares regression is that the sum of the squares of the errors is minimized. We can use calculus to find equations for the parameters $\beta_0$ and $\beta_1$ that minimize the sum of the squared errors, $S$.
$$S = \displaystyle\sum\limits_{i=1}^n \left(e_i \right)^2= \sum \left(y_i - \hat{y_i} \right)^2= \sum \left(y_i - \beta_0 - \beta_1x_i\right)^2$$
We want to find $\beta_0$ and $\beta_1$ that minimize the sum, $S$. We start by taking the partial derivative of $S$ with respect to $\beta_0$ and setting it to zero.
$$\frac{\partial{S}}{\partial{\beta_0}} = \sum 2\left(y_i - \beta_0 - \beta_1x_i\right)^1 (-1) = 0$$ $$\sum \left(y_i - \beta_0 - \beta_1x_i\right) = 0 $$ $$\sum \beta_0 = \sum y_i -\beta_1 \sum x_i $$ $$n\beta_0 = \sum y_i -\beta_1 \sum x_i $$ $$\beta_0 = \frac{1}{n}\sum y_i -\beta_1 \frac{1}{n}\sum x_i \tag{1}$$ $$\beta_0 = \bar y - \beta_1 \bar x \tag{*} $$
now take the partial of $S$ with respect to $\beta_1$ and set it to zero.
$$\frac{\partial{S}}{\partial{\beta_1}} = \sum 2\left(y_i - \beta_0 - \beta_1x_i\right)^1 (-x_i) = 0$$
$$\sum x_i \left(y_i - \beta_0 - \beta_1x_i\right) = 0$$
$$\sum x_iy_i - \beta_0 \sum x_i - \beta_1 \sum x_i^2 = 0 \tag{2}$$
substitute $(1)$ into $(2)$
$$\sum x_iy_i - \left( \frac{1}{n}\sum y_i -\beta_1 \frac{1}{n}\sum x_i\right) \sum x_i - \beta_1 \sum x_i^2 = 0 $$
$$\sum x_iy_i - \frac{1}{n} \sum x_i \sum y_i + \beta_1 \frac{1}{n} \left( \sum x_i \right) ^2 - \beta_1 \sum x_i^2 = 0 $$
$$\sum x_iy_i - \frac{1}{n} \sum x_i \sum y_i = - \beta_1 \frac{1}{n} \left( \sum x_i \right) ^2 + \beta_1 \sum x_i^2 $$ $$\sum x_iy_i - \frac{1}{n} \sum x_i \sum y_i = \beta_1 \left(\sum x_i^2 - \frac{1}{n} \left( \sum x_i \right) ^2 \right) $$ $$\beta_1 = \frac{\sum x_iy_i - \frac{1}{n} \sum x_i \sum y_i}{\sum x_i^2 - \frac{1}{n} \left( \sum x_i \right) ^2 } = \frac{cov(x,y)}{var(x)}\tag{*}$$
The partial derivatives of $$ Q(\alpha,\beta)=\sum_i (y_i-\alpha-\beta x_i)^2 $$ are $$ \frac{\partial Q}{\partial \alpha}(\alpha,\beta)=-2\sum_i(y_i-\alpha-\beta x_i)=-2(n\bar{y}-n\alpha-n\beta\bar{x})\tag{1} $$ and $$ \frac{\partial Q}{\partial \beta}(\alpha,\beta)=-2\sum_ix_i(y_i-\alpha-\beta x_i)=-2(\mathrm{SP}_{xy}-n\alpha\bar{x}-\beta\mathrm{SS}_x)\tag{2} $$ with $\mathrm{SP}_{xy}=\sum x_i y_i$ and $\mathrm{SS}_x=\sum x_i^2$. Putting $(1)$ equal to zero gives us $$ \hat{\alpha}=\bar{y}-\hat{\beta}\bar{x} $$ and plugging this into $(2)$ gives us the equation $$ \begin{align} 0=\mathrm{SP}_{xy}-n(\bar{y}-\hat{\beta}\bar{x})\bar{x}-\hat{\beta}\mathrm{SS}_x&=\mathrm{SP}_{xy}-n\bar{y}\bar{x}+n\hat{\beta}\bar{x}^2-\hat{\beta}\mathrm{SS}_x\\ &=\mathrm{SP}_{xy}-n\bar{y}\bar{x}-\hat{\beta}(\mathrm{SS}_x-n\bar{x}^2) \end{align} $$ and hence $$ \hat{\beta}=\frac{\mathrm{SP}_{xy}-n\bar{y}\bar{x}}{\mathrm{SS}_x-n\bar{x}^2}. $$
