I am been this equation but I can't follow how this holds.
Eq1:
$$e^{j\left(\omega -\omega_k\right)(N-1)(T/2)}\cdot\frac{(\sin((w-w_k)((NT/2)))}{(\sin((w-w_k)(T/2)}=\frac{1-e^{j\left(\omega -\omega_k\right)NT}}{1-e^{j\left(\omega -\omega_k\right)T}}$$
As I can see
Eq2:
$$e^{j\left(\omega -\omega_k\right)(N-1)(T/2)}={e^{j\left(\omega -\omega_k\right)NT/2}\cdot e^{-j\left(\omega -\omega_k\right)T/2}} = \frac{e^{j\left(\omega -\omega_k\right)NT/2}}{e^{j\left(\omega -\omega_k\right)T/2}}$$
We also know the identity that $\sin(\delta)= \dfrac{e^{j(\delta)}-e^{-j(\delta)}}{2j}$
Therefor I could use this identity to derive
Eq3
$$\frac{(\sin((w-w_k)((NT/2)))}{(\sin((w-w_k)(T/2)} = \frac{e^{j\left(\omega -\omega_k\right)NT/2}-e^{-j \left(\omega -\omega_k\right)NT/2}}{e^{j\left(\omega -\omega_k\right)T/2}-e^{-j\left(\omega -\omega_k\right)T/2}}$$
From this I get Eq2 times Eq3(right hand of equations)
$$\frac{e^{j\left(\omega -\omega_k\right)NT/2}}{e^{j\left(\omega -\omega_k\right)T/2}} \cdot \frac{e^{j\left(\omega -\omega_k\right)NT/2}-e^{-j \left(\omega -\omega_k\right)NT/2}}{e^{j\left(\omega -\omega_k \right)T/2}-e^{-j\left(\omega -\omega_k \right)T/2}} =$$
$$\frac{e^{j\left(\omega -\omega_k\right)NT}-1}{e^{j\left(\omega -\omega_k\right)T}-1}$$
Now this means that I have not been able to derive the result as stated in equation 1. And I must have done something wrong. Could someone please identify where in my steps I went wrong and how to derive the steps in between.
Hint:$$\frac{x-1}{y-1} = \frac{-(1-x)}{-(1-y)}=\frac{1-x}{1-y}$$