jmc

By the Pythagorean theorem, we have: $$\begin{array}{rl}A{C}^2& =A{B}^2+B{C}^2=1+1=2;\\ A{D}^2& =A{C}^2+C{D}^2=2+1=3;\\ A{E}^2& =A{D}^2+D{E}^2=3+1=4.\end{array}$$Thus $AE=\sqrt{4}=2,$ and the perimeter of pentagon $ABCDE$ is $1+1+1+1+2=\boxed{6}$.