Extending the result

Above we have shown that

$$I_n(x) = F_n(x) {\rm e}^x - (-1)^n n!$$

We now want to find

$$J := \int_0^x t^n {\rm e}^{at} {\rm d}t$$

All we need to do to get this into the form of $I_n$ is to substitute $u = at$ and ${\rm d}u = a{\rm d}t$,

$$J = \int_0^{ax} \frac{u^n}{a^n} {\rm e}^u \frac{{\rm d}u}{a} = \frac{1}{a^{n+1}} \int_0^{ax} u^n {\rm e}^u {\rm d}u =: \frac{1}{a^{n+1}} I_n(ax)$$

Now,

$$\frac{1}{a^{n+1}} I_n(ax) = {\rm e}^{ax} \sum_{r=0}^n (-1)^{n-r} \frac{n!}{r!} \frac{a^r x^r}{a^{n+1}} - \frac{(-1)^n n!}{a^{n+1}}$$

$$= {\rm e}^{ax} \sum_{r=0}^n (-1)^{n-r} \frac{n!}{r!} \frac{x^r}{a^{n-r+1}} - \frac{(-1)^n n!}{a^{n+1}}$$