How can you prove the Laplace transform of a periodic function with an integral?

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Here is this week's POTW:

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Suppose $f$ is a $p$-periodic complex-valued function on $[0,\infty)$. Let $F(s)$ denote the Laplace transform of $f(t)$. Prove

$$F(s) = \frac{1}{1 - e^{-ps}}\int_0^p e^{-st}f(t)\, dt, \qquad \operatorname{Re}(s) > 0.$$

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Congratulations to GJA, chisigma, and Kokuhaku for their correct solutions! Here is GJA's solution below.

Spoiler

We first note that, since $\text{Re}(s)>0$, $|e^{-sp}|=e^{-\text{Re}(s)p}<1$; therefore
$$\sum_{n=0}^{\infty}e^{-nsp}=\frac{1}{1-e^{-sp}}\tag{$*$}$$
via geometric series.

Using the definition of the Laplace Transform we compute:

$\begin{align*}
F(s)&=\int_{0}^{\infty}e^{-st}f(t)dt\\
&=\sum_{n=0}^{\infty}\int_{np}^{(n+1)p}e^{-st}f(t)dt\\
&=\sum_{n=0}^{\infty}\int_{0}^{p}e^{-s(u+np)}f(u+np)du\tag{$u=t-np$}\\
&=\sum_{n=0}^{\infty}e^{-nsp}\int_{0}^{p}e^{-su}f(u+np)du\\
&=\sum_{n=0}^{\infty}e^{-nsp}\int_{0}^{p}e^{-su}f(u)du\tag{$f$ is $p$-periodic}\\
&=\frac{1}{1-e^{-sp}}\int_{0}^{p}e^{-st}f(t)dt\tag{Via $(*)$ and setting $t=u$}
\end{align*}$

which gives the desired result.