Proving $\left(\frac{1+\sin x+i\cos x}{1+\sin x-i\cos x}\right)^n=\cos n\left(\frac{\pi }{2}-x\right)+i\sin n\left(\frac{\pi }{2-x}\right)$

How to solve the following question?

If $n$ is an integer, show that

\begin{eqnarray} \left(\frac{1+\sin x+i\cos x}{1+\sin x-i\cos x}\right)^n=\cos n\left(\frac{\pi }{2}-x\right)+i\sin n\left(\frac{\pi}{2}-x\right) \end{eqnarray}

where x is a real number such that $\left(1+\sin x\right)^2+\cos^2x>0$.

My work

I have no idea how to continue...

$$\frac{1+\sin x+i\cos x}{1+\sin x-i\cos x}=\frac{(1+\sin x+i\cos x)^2}{(1+\sin x)^2+\cos^2 x}$$

$$=\frac{2(1+\sin x)(\sin x+i\cos x)}{2(1+\sin x)}=\cos\left(\frac\pi2-x\right)+i\sin\left(\frac\pi2-x\right)$$

Apply de Movire's Theorem assuming $1+\sin x\ne0$

as $\sin x+1=0\implies 1+\sin x-i\cos x=0$