Least squares as an optimisation problem

Partial derivatives.
$$\frac{\partial F}{\partial m} = -2 \sum_{k=1}^n x_k(y_k - m x_k - p) = 0,$$
$$\frac{\partial F}{\partial p} = -2 \sum_{k=1}^n (y_k - m x_k - p) = 0.$$

Normal system. Dividing by $-2$:
$$\sum x_k y_k = m \sum x_k^2 + p \sum x_k, \qquad \sum y_k = m \sum x_k + n p.$$
From the second equation, $p = \bar y - m \bar x$. Substituting into the first:
$$\sum x_k y_k = m \sum x_k^2 + (\bar y - m \bar x) n \bar x = m \left( \sum x_k^2 - n \bar x^2 \right) + n \bar x \bar y.$$
Hence
$$m = \frac{\sum x_k y_k - n \bar x \bar y}{\sum x_k^2 - n \bar x^2} = \frac{\sum (x_k - \bar x)(y_k - \bar y)}{\sum (x_k - \bar x)^2}.$$
This is the slope of the least-squares line.

Global minimum. $F$ is a convex quadratic function (sum of squares of affine forms), so the critical point is a global minimum.