$\int_{0}^{a} \sin ^{\nu-2}(a-x) \sin (\nu x) d x=\frac{1}{\nu-1} \sin ^{\nu} a$ Prove that

$$I=\int_{0}^{a} \sin ^{\nu-2}(a-x) \sin (\nu x) d x=\frac{1}{\nu-1} \sin ^{\nu} a$$ if $a>0, \operatorname{Re}(\nu)>1$

Attempt:
Using the property $\int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx$ we get
$$I=\int_{0}^{a}\sin^{\nu-2}(x)\sin(\nu a-\nu x)\:dx$$
$\implies$
$$I=\sin(\nu a)\int_{0}^{a}\sin^{\nu -2}(x)\cos(\nu x)\:dx-\cos(\nu a)\int_{0}^{a}\sin^{\nu -2}(x)\sin(\nu x)dx$$
$\implies$
$$I=\sin(\nu a)I_1-\cos(\nu a)I_2$$
Where $$I_1=\int_{0}^{a}\sin^{\nu -2}(x)\cos(\nu x)\:dx$$ and
$$I_2=\int_{0}^{a}\sin^{\nu-2}(x)\sin(\nu x)\:dx$$
Now we have:
$$I_1+iI_2=\int_{0}^{a}\sin^{\nu -2}(x)e^{i\nu x}dx$$
Any way to proceed from here?
 A: $$\begin{align}
&\small\int_0^a\sin^{n-2}(a-x)\sin(nx)dx=\int_0^a\sin^{n-2}(t)\sin(n(a-t))dt\\
&\small=\int_0^a\left[\frac{e^{it}-e^{-it}}{2i}\right]^{n-2}\left[\frac{e^{in(a-t)}-e^{-in(a-t)}}{2i}\right]dt\\
&\small=\frac1{(2i)^{n-1}}\int_0^a \left[e^{it(n-2)}\left(1-e^{-2it}\right)^{n-2}\right]
\left[e^{in(a-t)}-e^{-in(a-t)}\right]dt\\
&\small=\frac1{(2i)^{n-1}}\int_0^a \left[e^{it(n-2)}\sum_{k\ge0}(-1)^k\binom{n-2}{k}e^{-2itk}\right]
\left[e^{in(a-t)}-e^{-in(a-t)}\right]dt\\
&\small=\frac1{(2i)^{n-1}}\sum_{k\ge0}(-1)^k\binom{n-2}{k}\left\{e^{ina}\int_0^ae^{-2it(k+1)}dt-e^{-ina}\int_0^ae^{2it(n-k-1)}dt\right\}\\
&\small=\frac1{(2i)^{n-1}}\sum_{k\ge0}(-1)^k\binom{n-2}{k}\left\{e^{ina}\left[\frac{e^{-2it(k+1)}}{-2i(k+1)}\right]_0^a-e^{-ina}\left[\frac{e^{2it(n-k-1)}}{2i(n-k-1)}\right]_0^a\right\}\\
&\small=\frac{-1}{(2i)^{n}}\sum_{k\ge0}(-1)^k\binom{n-2}{k}\left\{e^{ina}\frac{e^{-2ia(k+1)}-1}{k+1}+e^{-ina}\frac{e^{2ia(n-k-1)}-1}{n-k-1}\right\}\\
&\small=\frac{-1}{(2i)^{n}}\sum_{k\ge0}(-1)^k\binom{n-2}{k}\left\{
\frac{n\, e^{ia(n-2k-2)}}{(k+1)(n-k-1)}-\frac{e^{ina}}{k+1}-\frac{e^{-ina}}{n-k-1}\right\}\\
&\small=\frac{1}{(2i)^{n}}\left\{\frac{e^{ina}}{n-1}\sum_{k\ge0}(-1)^{k+1}\binom{n}{k+1}e^{-2ia(k+1)}+\sum_{k\ge0}(-1)^k\binom{n-2}k\left[\frac{e^{ina}}{k+1}+\frac{e^{-ina}}{n-k-1}\right]\right\}\\
&\small\stackrel{(*)}=\frac{1}{(2i)^{n}}\left[\frac{(e^{ia}-e^{-ia})^n-e^{ina}}{n-1}+\frac{e^{ina}}{n-1}\right]\\
&\small=\frac{\sin^n a}{n-1},
\end{align}$$
where in $(*)$ we used:
$$\begin{align}
&\small\sum_{k\ge0}(-1)^{k+1}\binom{n}{k+1}e^{-2ia(k+1)}=\left(1-e^{-2ia}\right)^n-1\\
&\small\sum_{k\ge0}(-1)^k\binom{n-2}k\frac1{k+1}=\frac{-1}{n-1}\sum_{k\ge0}(-1)^{k+1}\binom{n-1}{k+1}=\frac{1}{n-1},\\
&\small\sum_{k\ge0}(-1)^k\binom{n-2}k\frac1{n-k-1}=\frac{1}{n-1}\sum_{k\ge0}(-1)^{k}\binom{n-1}{k}=0.
\end{align}$$
