Proving the Laplace transform of $t^n e^{at}$ using mathematical induction Given the problem of finding the Laplace transform of the function $$f(t)=t^ne^{at}$$ with $n\in\mathbb{N}$ and $a\in\mathbb{R}$, I realize it can be shown the transform is $$\frac{n!}{(s-a)^{n+1}}$$ by more than one direct method. However, I'd like to show this using strong induction. I began the problem by showing that
\begin{align*}
\mathcal{L}\{te^{at}\}=\int_{0}^{\infty}te^{-(s-a)t}\,dt&=\frac{1}{(s-a)^2},\\
\mathcal{L}\{t^2e^{at}\}=\int_{0}^{\infty}t^2e^{-(s-a)t}\,dt&=\frac{2}{(s-a)^3},\text{ and}\\
\mathcal{L}\{t^3e^{at}\}=\int_{0}^{\infty}t^3e^{-(s-a)t}\,dt&=\frac{6}{(s-a)^4}.
\end{align*}
I originally did this so I could personally see the pattern. Then, I began the proof as follows.
$\textbf{Proof:}$ Let $P(n)$ be the statement
$$P(n): \mathcal{L}\{t^ne^{at}\}=\frac{n!}{(s-a)^{n+1}},$$
for all $n\in\mathbb{N}.$ We have already shown $P(1), P(2), P(3)$ are true, so the base case has been proven.
Inductive Step: Assume $P(k)$ is true for all $k$. We must then show that $P(k)\implies P(k+1).$
We see
\begin{align*}
\mathcal{L}\{t^{k+1}e^{at}\}&=\frac{(k+1)!}{(s-a)^{k+2}}\\
&=\frac{k+1}{s-a}\frac{k!}{(s-a)^{k+1}}\\
&=\frac{k+1}{s-a}\mathcal{L}\{t^{k}e^{at}\} \quad\text{(by the inductive step)}
\end{align*}
At this point, I'm not really sure where to go. I'm pretty awful at proofs, so I assume I'm probably not going in the right direction after the inductive hypothesis.
 A: You can usethe integral definition of the Laplace transform:
$$\mathcal{L}\{t^{n+1}e^{at}\}=\int_0^{\infty}t^{n+1}e^{-t(s-a)} dt$$
Integrate by part and  use the reccurence for the second integral.
$$\mathcal{L}\{t^{n+1}e^{at}\}= \lim_{N\to\infty} \left[t^{n+1}\dfrac {e^{-t(s-a)} }{a-s}\right]_0^{N}- \dfrac {n+1}{a-s}\mathcal{L}\{t^{n}e^{at} \}$$
A: My attempt :
$$ \mathcal{L}\{t^ne^{at}\}= \frac{n!}{(s-a)^{n+1}}$$
For $P(k):$
$$ \tag{1}\mathcal{L}\{t^ke^{at}\}= \frac{k!}{(s-a)^{k+1}}$$
Need to prove for $P(k+1) :$
$$\tag{2}\mathcal{L}\{t^{k+1}e^{at}\}= \frac{(k+1)!}{(s-a)^{k+2}}$$
Step 1 :
$$ \mathcal{L}\{t^{k+1}e^{at}\}=\int^{\infty}_{0} \left (t^{k+1}e^{at} \right)e^{-st}dt $$
$$\mathcal{L}\{t^{k+1}e^{at}\}=\int^{\infty}_{0} \left (t^{k+1}e^{-t(s-a)} \right)dt $$
Using integration by part :
$$ \int udv= uv -\int v du$$
$u = t^{k+1}$
$ \frac{du}{dt}=(k+1)~t^k$
$ \frac{dv}{dt}= e^{-t(s-a)}$
$ v = - \frac{1}{s-a}~e^{-t(s-a)}$
So :
Note :
$$\mathcal{L}\{t^ke^{at}\}=\int^{\infty}_{0} t^{k} e^{-t(s-a)}dt$$
$$\mathcal{L}\{t^{k+1}e^{at}\}=  \lim_{N\to\infty} \left [- \frac{t^{k+1}}{s-a}e^{-t(s-a)} \right]^{N}_{0} + \frac{k+1}{s-a} \int^{\infty}_{0}t^{k}e^{-t(s-a)}dt$$
$$\mathcal{L}\{t^{k+1}e^{at}\}=0+0+\frac{k+1}{s-a}~ \mathcal{L}\{t^ke^{at}\} $$
$$\mathcal{L}\{t^{k+1}e^{at}\}= \frac{k+1}{s-a} \frac{k!}{(s-a)^{k+1}}$$
Final step, proving (2):
$$\boxed{\mathcal{L}\{t^{k+1}e^{at}\}= \frac{(k+1)!}{(s-a)^{k+2}}}$$
