My attempt:
Let $$u=\sin^nx \rightarrow \text{d}u=n\sin^{n-1}x\cos x\text{d}x $$ and $$ v = \int{e^{ax}}\text{d}x=\frac{e^{ax}}{a}.$$
Now we have $u\cdot v- \int{v\cdot \text{d}u} $, so
$$I_n = \frac{e^{ax}}{a}\sin^nx-\frac{n}{a}\int{e^{ax}\sin^{n-1}x\cos x\text{d}x}$$
Now solving: $\int{e^{ax}\sin^{n-1}x\cos x\text{d}x}$, Let $$ u = e^{ax} \rightarrow \text{d}u=ae^{ax}$$ and $$ v = \int{\sin^{n-1}x \cos x \text{d}x} = \frac{\sin^nx}{n}. $$ So, $$\int{e^{ax}\sin^{n-1}x\cos x\text{d}x} = \frac{\sin^nx}{n}e^{ax}-\frac{a}{n}I_n.$$ We win $$I_n = \frac{e^{ax}}{a}\sin^nx-\frac{n}{a}\left(\frac{\sin^nx}{n}e^{ax}-\frac{a}{n}I_n\right)$$ $$ 0 = 0$$
Hint. We assume $a \neq0, \pm i$. From the identity you have already obtained by parts, $$ I_n = \frac{e^{ax}}{a}\sin^nx-\frac{n}{a}\int{e^{ax}\sin^{n-1}x\cos x\text{d}x} \tag1 $$ you may rather perform another integration by parts as follows $$ \begin{align} &\int{e^{ax}\sin^{n-1}x\cos x\text{d}x} \\&= \frac{e^{ax}}{a}\sin^{n-1}x\cos x-\frac{1}{a}\int{e^{ax}\left[(n-1)\sin^{n-2}x\cos^2 x-\sin^{n}x\right]\text{d}x} \tag2 \end{align} $$ replacing $\cos^2 x$ with $1-\sin^2 x$ then inserting $(2)$ in $(1)$, one gets $$ I_n=\frac{e^{ax}}{a}\sin^{n}x-\frac{ne^{ax}}{a^2}\sin^{n-1}x\cos x+\frac{n(n-1)}{a^2}I_{n-2}-\frac{n^2}{a^2}I_{n} $$ or
with $I_0=\dfrac{e^{ax}}a$ and $I_1=\dfrac{-\cos x+a \sin x}{1+a^2}e^{a x}$.