The Fourier of a sum is the sum of the Fouriers; that is what makes the Fourier transform a linear operation. Thus, $$\mathcal{F}[\Sigma_{n\in\mathbb{Z}}x(n)\delta(t-nt_0)] = \Sigma_{n\in\mathbb{Z}}x(n)e^{-2\pi iknt_0} $$
In other words, the Fourier transform of a series of modulated Dirac deltas is the sum of the transforms of the individual deltas.
(I changed the notation from $\delta(t-nt)$ to $\delta(t-nt_0)$ since I'm assuming that as $n$ increases we're going forward in time in discrete steps, and not moving around inside a single Dirac delta's zero zone, changing the $n$ of $\delta(t\cdot(1-n))$.)
Let $f$ be a smooth integrable function as you commented. Then we can define the Fourier Transform as
$$\mathcal{F}[f](y) := \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f(x)e^{-iyx}\mathrm{d}x$$
We can also derive the inverse of Fourier Transform as:
$$\mathcal{F}^c[f](y):=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f(x)e^{+iyx}\mathrm{d}x$$
Now lets see what happen wen we apply both to a function
$$f(y) = \mathcal{F}^c[\mathcal{F}[f]](y) = \frac{1}{2\pi}\int_\mathbb{R}\left(\int_{\mathbb{R}} f(\omega)e^{-i\omega x}\mathrm{d}\omega \right)e^{+iyx}\mathrm{d}x = \int_{\mathbb{R}}f(\omega)\left(\int_{\mathbb{R}}\frac{1}{2\pi}e^{-i(\omega-y)x}\mathrm{d}x\right)\mathrm{d}\omega$$
We can see in this last equality that the function in brakets acts as a Dirac Delta. So we can relate this as de Dirac as a notation (because the Dirac Delta just have formal sense in distribution theory)
$$\delta(\omega-y) = \frac{1}{2\pi}\int_{-\infty}^{+\infty}e^{-i(\omega-y)x}\mathrm{d}x$$
This is an intuitive constructive way of seen that the Dirac is related to Fourier Transforms. Now for your question we can use the definition on how an Dirac Delta acts
$\hat{\delta}(\omega) = \frac{1}{(2\pi)^{1/2}}\int \delta(t-t_0)e^{-i\omega t}\mathrm{d}t = \frac{1}{(2\pi)^{1/2}}e^{-\omega t_0}$
Where we just used the above
$$f(y) = \int f (\omega)\delta(\omega-y)\mathrm{d}\omega$$
For $f(t):=e^{-i\omega t}/\sqrt{2\pi}$
Best Answer
You can view this as a limiting process. Start with the truncated integral: $$ \frac{1}{2\pi}\int_{R}^{R}e^{j\omega t}dt = \frac{1}{\pi}\frac{\sin(R\omega)}{\omega} $$ If you integrate this against a function and take the limit as $R\rightarrow\infty$, then $$ \lim_{R\rightarrow\infty}\int_{-\infty}^{\infty}f(\omega)\frac{1}{2\pi}\int_{-R}^{R}e^{j\omega t}dt d\omega = \lim_{R\rightarrow\infty}\frac{1}{2\pi}\int_{-R}^{R}\int_{-\infty}^{\infty}f(\omega)e^{j\omega t}d\omega dt. $$ The above is $(f^{\wedge})^{\vee}(0)=f(0)$ if $f$ has some smoothness at $0$.