Unbelievably simple

$\displaystyle I = \int_{0}^1 (ln x)^{100} dx =$ $(x = e^{t}, dx = e^{t} dt)$ $= \int_{-\infty}^0 t^{100} \cdot e^{t} dt =$ (Integrating by parts) $= \left[t^{100} e^{t} \right]_{-\infty}^0 - 100 \int_{-\infty}^0 t^{99}e^{t} dt = -100 \cdot (-99)\int_{-\infty}^0 t^{98} et dt =$ $= (-1)^{99}100! \int_{-\infty}^0 t e^{t} dt = (-1)^{100}100! \int_{-\infty}^0 e^{t} dt = 100! \Rightarrow$ the number of trailing zeroes of I = $\frac{100}{5} + \frac{100}{25} = \boxed{24}$ or $\int_{-\infty}^0 t^{100} \cdot e^{t} dt =$ $(t = - u; dt = -du)$ $= \int_{0}^\infty u^{100} \cdot e^{-u} du = \Gamma(101) = 100!$