Determine the Laplace Transform $(t-1)^4$ How do I find the Laplace transform of 
$ (t-1)^{4} $?
by using properties and a table of common Laplace transforms? 
I know I could just turn the equation into a polynomial and use $ t^{n} \to n!/(s^{n+1}) $ , but I want to know if there is a way to do this without turning the equation into a polynomial.
 A: Hint: First write $ (t-1)^{4}=t^4-4t^3+6t^2-4t+1$ and then find the Laplace transform of each term by using the formula you mention above. Or you can directly by using the integral formula for the Laplace transform by substituting and integrating four times together with the use of integration by parts..
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
&\color{#0000ff}{\large\int_{0}^{\infty}\pars{t - 1}^{4}\expo{-st}\,\dd t}
=
\expo{-s}\int_{-1}^{\infty}t^{4}\expo{-st}\,\dd t
=
\int_{-s}^{\infty}t^{4}\expo{-st}\,\dd t
=
{\expo{-s} \over s^{5}}\,\Gamma\pars{5,-s}
\\[3mm]&=
{\expo{-s} \over s^{5}}\bracks{\pars{5 - 1}!\expo{-\pars{-s}}
\sum_{k = 0}^{5 - 1}{\pars{-s}^{k} \over k!}}
=
24\sum_{k = 0}^{4}\pars{-1}^{k}\,{s^{k - 5} \over k!}
=
\color{#0000ff}{\large%
{24 \over s^{5}} - {24 \over s^{4}} + {12 \over s^{3}} - {4 \over s^{2}}
+
{1 \over s}}
\end{align}
$\Gamma\pars{s,x}$ is the $\tt\mbox{Upper Incomplete Gamma function}$.
A: Because:
$$\mathscr{L}\{t^4\}=\frac{4!}{s^5}$$
Using the time translation property:
$$\mathscr{L}\{f(t-a)\}=e^{-a}F(s)$$
You immediately obtain that:
$$\mathscr{L}\{(t-1)^4\}=\frac{4!}{s^5}e^{-s}$$ 
