Non-uniqueness of continuous time sinusoids? I have read the DSP literature and I get why:
$$e^{j(\omega+2 \pi m)n}=e^{j\omega n}$$

But can't we prove the same in continuous domain also in the following way:
$$e^{j\omega t}$$
$$e^{j(\omega+2 \pi m)t}$$
$$\left( e^{j\omega} \cdot e^{2j\pi m}\right) e^t$$
$$e^{j\omega t}$$

If my above reasoning is correct, then does not it mean that analog frequencies separated by $2\pi$ are same? But I know this is not the case.

Please help me in this case.
In discrete-time we have

$$e^{j\omega_0n}=e^{j(\omega_0+2\pi k)n},\quad \forall n$$

because $2\pi kn$ is an integer multiple of $2\pi$ if $k$ and $n$ are integers, and, consequently, $e^{j2\pi kn}=1$. In continuous time you generally have

$$e^{j\omega_0t}\neq e^{j(\omega_0+2\pi k)t}$$

because $2\pi k t$ is (for arbitrary $t$) not an integer.