As it says in the title, I would like to show that $e^{\alpha t} \ast (\frac{t^k}{k!}e^{\alpha t}) = \frac{t^{k+1}}{(k + 1)!}e^{\alpha t}$ for the convolution. However the only thing I know for the convolution is the convolution of two function $f,g$ defined by:
$$(f \ast g)(t) := \int\limits_0^tf(t - s)g(s)\,\mathrm{d}s$$
Since the exponential function is not any function, I was wondering if there is a special way to represent it for the convolution (in terms of integral between $0$ and $t$) or if I could just replace $f$ and $g$ by $e^{\alpha t}$ and $(\frac{t^k}{k!}e^{\alpha t})$ respectively?
This stops me from showing the equality itself, so any clue would be welcome!
By definition $f(t)=e^{\alpha t}$ and $g(t)=\frac{t^k}{k!}e^{\alpha t}$. Plugging these function into the integral, $$\int\limits_0^tf(t - s)g(s)ds=\int\limits_0^t e^{\alpha(t-s)}\cdot\frac{s^k}{k!}e^{\alpha s}ds.$$ We may pull out the constants (in terms of $s$) to get to$$\frac{e^{\alpha t}} {k!}\int\limits_0^t s^k\underbrace{e^{-\alpha s}\cdot e^{\alpha s}}_{=1} ds=\frac{e^{\alpha t}} {k!}\int\limits_0^t s^k ds=\frac{e^{\alpha t}}{k!}\left[\frac{t^{k+1}}{k+1}-\frac{0^{k+1}}{k+1}\right]=\frac{t^{k+1}}{(k + 1)!}e^{\alpha t}.$$ In the last step we used the defining property $k!\cdot(k+1)=(k+1)!$ of the factorial.