I have tried to find the Laplace transform of $L_n(t)$, which is the $n$th order Laguerre polynomial defined by $$L_n(t) :=e^t \frac{\mathrm{d}^n}{\mathrm{d}t^n}(t^n e^{-t}).$$
I have managed to get
$$\boxed{\cal{L}_n(p)=\dfrac{(p-1)^n\cdot n!}{p^{n+1}}}$$
and I understand how to use all the appropriate properties of the Laplace transform; in particular, that
$$\cal{L}\left[\frac{\mathrm{d}^n}{\mathrm{d}t^n}(t^n e^{-t})\right]=p^n\cal{L}[t^n e^{-t}]-p^{n-1}\left[\frac{\mathrm{d}}{\mathrm{d}t}(t^n e^{-t})\right]_{t=0}-p^{n-2}\left[\frac{\mathrm{d}^{2}}{\mathrm{d}t^{2}}(t^n e^{-t})\right]_{t=0}-\ldots -p\left[\frac{\mathrm{d}^{n-2}}{\mathrm{d}t^{n-2}}(t^n e^{-t})\right]_{t=0}-\left[\frac{\mathrm{d}^{n-1}}{\mathrm{d}t^{n-1}}(t^n e^{-t})\right]_{t=0}$$
and all but the first term vanish. Then
$$\begin{align} \cal{L}[t^n e^{-t}]&=(-1)^n\frac{\mathrm{d}^n}{\mathrm{d}p^n}(\cal{L}[e^{-t}])\\ & =(-1)^n \frac{\mathrm{d}^n}{\mathrm{d}p^n}\left(\frac{1}{p+1}\right)\\ &=(-1)^n\left[(-1)^n\frac{n!}{(p+1)^n}\right]\\ &=\frac{n!}{(p+1)^n}\\ \end{align}$$
However, the book I am reading (Differential and Integral Equations by P.J. Collins) shows the same answer as I got (boxed above), but missing the $n!$ on top of the answer. Is this an error in the book or have I gone wrong somewhere?