Laplace Transforms · Hard problem

Let $g(t) = \displaystyle\int_0^t \tau e^{-\tau}\,d\tau$. Using $\mathcal{L}\left\{\int_0^t f(\tau)\,d\tau\right\} = \dfrac{F(s)}{s}$, find $\mathcal{L}\{g\}$ evaluated at $s = 1$.