Is there a simple algebraic proof?
Let the glide reflection be $T$. Without loss of generality the reflection part of $T$ is reflection in the $y$-axis. Assume that the translation part is by the vector $(a,b)$.
The reflection takes $(x,y)$ to $(-x,y)$. The translation part now takes us to $(-x+a,y+b)$. So $T$ takes $(x,y)$ to $(-x+a,y+b)$.
Do it again. The reflection part takes $(-x+a,y+b)$ to $(x-a,y+b)$, and the translation part takes this to $(x,y+2b)$. So $T^2$ is translation parallel to the $y$-axis by $2b$.