# Use the definition of Laplace Transforms to show that:displaystyle{L}{leftlbrace{t}^{n}rightrbrace}=frac{{{n}!}}{{{s}^{{{n}+{1}}}}},{n}={1},{2},{3},ldots

Use the definition of Laplace Transforms to show that:
$L\left\{{t}^{n}\right\}=\frac{n!}{{s}^{n+1}},n=1,2,3,\dots$

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Step 1
Given:
The laplace transformation,
$L\left\{{t}^{n}\right\}=\frac{n!}{{s}^{n+1}}$
Step 2
Proof:
Consider $L\left({t}^{n}\right)$ and simplify it as follows.
$L\left\{f\left(t\right)\right\}={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$
$L\left\{{t}^{n}\right\}={\int }_{0}^{\mathrm{\infty }}{e}^{-st}\left({t}^{n}\right)dt$
Apply integration by parts, $uv=uv-intuv$
Here,
$u={t}^{n}dv={e}^{-st}$

$L\left\{{t}^{n}\right\}={\left[\frac{-{t}^{n}\cdot {e}^{-st}}{s}\right]}_{0}^{\mathrm{\infty }}-{\int }_{0}^{\mathrm{\infty }}n{t}^{n-1}\left(\frac{{e}^{-st}}{s}\right)dt$
$=0+\frac{n}{s}{\int }_{0}^{\mathrm{\infty }}{t}^{n-1}\left({e}^{-st}\right)dt$
$=\frac{n}{s}L\left({t}^{n-1}\right)$
Step 3
Now substitute the values 1, 2, 3... for n.
When n=1
$L\left\{{t}^{1}\right\}=\frac{1}{s}L\left(1\right)$
$=\frac{1}{s}\left(\frac{1}{s}\right)$
$=\frac{1}{{s}^{2}}$
When n=2
$L\left\{{t}^{2}\right\}=\frac{2}{s}\left(L\left({t}^{1}\right)\right)$
$=\frac{2}{s}\left(\frac{1}{{s}^{2}}\right)$
$=\frac{2!}{{s}^{3}}$
Step 4
When n = 3,
$L\left\{{t}^{3}\right\}=\frac{3}{s}L\left({t}^{3-1}\right)$
$=\frac{3}{s}\left(L\left({t}^{2}\right)\right)$
$=\frac{3}{s}\left(\frac{2}{{s}^{3}}\right)$
$=\frac{6}{{s}^{3+1}}$
$=\frac{3!}{{s}^{3+1}}$
Step 5
Thus, for n = n,
$L\left({t}^{n}\right)=\frac{n!}{{s}^{n+1}}$
Hence it is proved.