The residuals $\boldsymbol{\epsilon}$ are in turn given by

$$\boldsymbol{\epsilon} = \boldsymbol{y}-\boldsymbol{\tilde{y}} = \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta},$$

and with

$$\boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)= 0,$$

we have

$$\boldsymbol{X}^T\boldsymbol{\epsilon}=\boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)= 0,$$

meaning that the solution for $\boldsymbol{\beta}$ is the one which minimizes the residuals.