= Solution
The convolution is
$$
(f*g)(x)=\int_{-\infty}^{\infty}f(x-y)g(y)\,dy.
$$
We claim that the $n$-fold convolution is
$$
\boxed{
F_n(x)=
\begin{cases}
\dfrac{x^{n-1}}{(n-1)!}e^{-x},&x\geq0,\\
0,&x<0.
\end{cases}}
$$
This is true for $n=1$. If it holds for $n-1$ and $x\geq0$, then
$$
\begin{aligned}
F_n(x)
&=\int_0^x e^{-(x-y)}
\frac{y^{n-2}e^{-y}}{(n-2)!}\,dy\\
&=\frac{e^{-x}}{(n-2)!}\int_0^xy^{n-2}\,dy
=\frac{x^{n-1}e^{-x}}{(n-1)!},
\end{aligned}
$$
and the convolution vanishes for $x<0$. This is the <gamma density from repeated exponential convolution>.
The Fourier transform is
$$
\boxed{
\widehat F_n(k)=\frac1{(n-1)!}
\int_0^\infty x^{n-1}e^{-(1+ik)x}\,dx
=\frac1{(1+ik)^n}}.
$$
The <convolution theorem> states
$$
\boxed{\widehat{f*g}(k)=\widehat f(k)\widehat g(k)}.
$$
Indeed, Fubini and $u=x-y$ give
$$
\begin{aligned}
\widehat{f*g}(k)
&=\int\!\int e^{-ikx}f(x-y)g(y)\,dy\,dx\\
&=\int\!\int e^{-ik(u+y)}f(u)g(y)\,du\,dy
=\widehat f(k)\widehat g(k).
\end{aligned}
$$
Since $\widehat F(k)=(1+ik)^{-1}$, induction immediately verifies
$$
\widehat F_n=(\widehat F)^n=(1+ik)^{-n}.
$$
For <Parseval identity>, take $g(x)=\overline{f(-x)}$. Then
$$
(f*g)(0)=\int_{-\infty}^{\infty}|f(x)|^2\,dx,
\qquad
\widehat g(k)=\overline{\widehat f(k)}.
$$
Using Fourier inversion at zero, which follows from the supplied delta identity,
$$
(f*g)(0)=\frac1{2\pi}\int_{-\infty}^{\infty}
\widehat{f*g}(k)\,dk,
$$
and the convolution theorem gives
$$
\boxed{
\int_{-\infty}^{\infty}|f(x)|^2\,dx
=\frac1{2\pi}\int_{-\infty}^{\infty}|\widehat f(k)|^2\,dk}.
$$
Apply this to $F_{n+1}$. Since
$$
\int_0^\infty|F_{n+1}(x)|^2dx
=\frac1{(n!)^2}\int_0^\infty x^{2n}e^{-2x}\,dx
=\frac{(2n)!}{2^{2n+1}(n!)^2},
$$
one obtains the <Even rational Parseval integral>
$$
\boxed{
\int_{-\infty}^{\infty}\frac{dk}{(1+k^2)^{n+1}}
=\frac{\pi(2n)!}{2^{2n}(n!)^2}
=\frac\pi{4^n}\binom{2n}{n}}.
$$
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