$2pi$ periodicity of discrete-time Fourier transform

In my signals and systems course, we have learned that the discrete-time Fourier transform is $2\pi$ periodic, but the continuous-time Fourier transform is not periodic in general. For reference, we are using the following definitions of each transform:

Continuous-Time:

$$ x(t) = \int_{-\infty}^{\infty} X(j\omega) e^{j\omega t} \, d\omega \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ X(j\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} \, dt $$

Discrete-Time:

$$ x[n] = \int_{\langle 2\pi \rangle} X(e^{j\omega}) e^{j\omega n} \, d\omega \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} x[n] e^{-j\omega n} $$

I'm searching for some intuition as for why the DTFT is periodic, but the CTFT is not. In class, my instructor presented the following argument: for a discrete-time signal,

$$ e^{j\omega n} = e^{j\omega(n + 2\pi)} $$

and thus any $x[n]$ can be expressed as a sum of individually $2\pi$ periodic functions. However, I don't see why that argument only applies to discrete-time signals - I feel as if it also works for continuous-time signals.

Any someone explain?