Why does the Laplace transform of a matrix exponential $f(t) = e^{At}$ satisfy $sF(s) = AF(s) + I$ Where $A$ is some $n \times n$ matrix 
Suppose I am given $f(t) = e^{At}$, then $\dot f(t) = Ae^{At}$, so $L(\dot f(t)) = sF(s) = AF(s)$
Why does $sF(s) = AF(s) + I$ hold true. (more specifically, where does the $I$ come from?)
 A: Let's take the scalar case for simplicity. $f(t)=e^{at}$ and $a<0$, so
$$
F(s)=\int_0^\infty e^{at}e^{-st}\,dt=\left[\frac{e^{(a-s)t}}{a-s}\right]_0^\infty=\frac{1}{s-a}, \qquad \text{Re}\,s\ge 0.
$$
Now
$$
sF(s)=\frac{s}{s-a}=\frac{s-a+a}{s-a}=1+\frac{a}{s-a}=1+aF(s).
$$
What this $1$ came from? It comes from $f(0)$: $L(\dot f)=sF(s)-f(0)$. For matrices is just the same.
A: We can write $f(t)=e^{At}$ as 
$$e^{At}=\sum_{n=0}^{\infty}\frac{t^nA^n}{n!}$$
Then, assuming that $||A||<s, $the Laplace Transform of $e^{At}$ becomes
$$\begin{align}
F(s)&=\mathscr{L}\left(e^{At}\right)(s)\\\\
&=\int_0^{\infty} e^{At}e^{-st}dt\\\\
&=\sum_{n=0}^{\infty}\frac{A^n}{n!}\int_0^{\infty}t^ne^{-st}dt\\\\
&=\sum_{n=0}^{\infty}\frac{A^n}{n!}\frac{\Gamma(n+1))}{s^{n+1}}\\\\
&=\sum_{n=0}^{\infty}\frac{A^n}{s^{n+1}}\\\\
&=\frac1s\left(I-\frac1sA\right)^{-1} \tag 1
\end{align}$$
Now, multiplying both sides of $(1)$ by $\left(I-\frac1sA\right)$ reveals
$$\left(I-\frac1sA\right)F(s)=I$$
whereupon rearranging terms and recognizing that $IF(s)=F(s)$, we obtain
$$\bbox[5px,border:2px solid #C0A000]{sF(s)=AF(s)+I}$$
as expected!
