Using the given identities, find the inverse DTFT
As pointed out in the comments, it is important for you to know and understand that multiplication in one domain (time or frequency) corresponds to convolution in the other domain. And, as a consequence, multiplication with $e^{-j\omega}$ in the Fourier domain corresponds to a delay of one sample in the time domain (i.e., convolution with $\delta[n-1]$). So you have

$$\begin{align}\mathcal{F}^{-1}\left\{B\frac{e^{-j\omega}}{1-ae^{-j\omega}}\right\}[n]&amp;=B\cdot\mathcal{F}^{-1}\left\{\frac{e^{-j\omega}}{1-ae^{-j\omega}}\right\}[n]\quad (\textrm{due to linearity of }\mathcal{F}^{-1})\\&amp;=B\cdot\mathcal{F}^{-1}\left\{\frac{1}{1-ae^{-j\omega}}\right\}[n-1]\\&amp;=B\cdot a^{n-1}u[n-1]\end{align}$$

where I used $\mathcal{F}^{-1}$ to denote the inverse discrete-time Fourier transform.