Why is $\int^\infty _{0^-}\delta(t-nT)e^{-st}dt = e^{-nsT}$?
Remember the definition of the delta function: it integrates $1$ all over the t-axis but it is zero for any $t\neq 0$. This means that if we multiply any function by $\delta (t)$, then the function is going to be zero everywhere except for $t=0$. In this case, we have

$$ \delta (t-nT)e^{-st}=
\left\{
 \begin{array}{ll}
  \delta (t-nT)e^{-snT}  &amp; \mbox{if } t = nT \\
  0 &amp; \mbox{if } t\neq nT
 \end{array}
\right.$$

So if we integrate this function, we have that the integral is zero for every point except for $t=nT$. So

$$\int_0^\infty \delta (t-nT)e^{-st}dt=\int_0^\infty \delta (t-nT)e^{-snT}dt=e^{-snT}\int_0^\infty \delta (t-nT)dt$$

But by definition, $$\int_0^\infty \delta (t-nT)dt=\int_{-\infty}^\infty \delta (t-nT)dt=1$$

So $$\int_0^\infty \delta (t-nT)e^{-st}dt=e^{-snT}$$

Sidenote: For this to be true, $n$ must be positive. If it is negative, then the integral will be $0$ (as the function inside the integral will be zero for all the integration interval).