Continuous-time vs Discrete-time Fourier transform
This is pretty straightforward to show. Let $x[n]$ be a sampled version of a continuous-time signal $x_c(t)$:

$$x[n]=x_c(nT)\tag{1}$$

The DTFT of $x[n]$ is defined by

$$X(e^{j\omega})=\sum_{n=-\infty}^{\infty}x[n]e^{-jn\omega}\tag{2}$$

The CTFT of the signal

$$x_c(t)\cdot\sum_{n=-\infty}^{\infty}\delta(t-nT)=\sum_{n=-\infty}^{\infty}x_c(nT)\delta(t-nT)\tag{3}$$

is given by

$$\begin{align}\mathscr{F}\left\{\sum_{n=-\infty}^{\infty}x_c(nT)\delta(t-nT)\right\}&amp;=\sum_{n=-\infty}^{\infty}x_c(nT)\mathscr{F}\left\{\delta(t-nT)\right\}\\&amp;=\sum_{n=-\infty}^{\infty}x_c(nT)e^{-jn\Omega T}\tag{4}\end{align}$$

which equals $(2)$ with $x[n]=x_c(nT)$ and $\omega=\Omega T$ being the normalized angular frequency.

And now I found this question, which is pretty similar but also asks about some background.