@MattL suggested a nice, simple way to see the above result.
But if you want to see the result in the normal analysis equations you mentioned, you can do like below.
Say S(t) is a periodic train of impulses.So S(t) can be written as
$$\ S(t)= \sum_{n=-\infty}^{\infty} \delta(t-nT)$$
Now if you take the fourier series of S(t),you can write S(t) as
$$S(t) =\sum_{n=-\infty}^{\infty} C_ke^{jnw_ot} $$
Where $C_n$ are exponential fourier series coefficients and $w_o$ is the fundamental frequency .
So from exponential fourier series we know that
$$C_n= (1/T)\int_{-T/2}^{T/2} S(t)e^{-jnw_ot} dt$$
Now in the above expression substitute the value of S(t) from the first expression.
So $$C_n = (1/T)\sum_{n=-\infty}^{\infty}\int_{-T/2}^{T/2} \delta(t-nT)e^{-jnw_ot} dt$$
Now, you have to make an observation, if you observe the integral, it's from -T/2 to +T/2. During this integral period, observe that only a single impulse $\delta(t)$ exists.All the other impulse functions in the summation occur after T/2 or before -T/2. So in total the above equation for $C_n$ can be written as
$$C_n = (1/T)\sum_{n=-\infty}^{\infty} \delta(t)e^{-jnw_ot}$$
From sifting property we can write the above as
$$C_n = (1/T)e^{-jw_on(0)} = (1/T)$$
Now put this value of $C_n$ in the first S(t) equation
$$S(t) = (1/T)\sum_{n=-\infty}^{\infty} e^{jnw_ot}$$
Now find the fourier transform of above equation
$$1\Longleftrightarrow 2\pi\delta(w)$$
$$e^{jw_ot}\Longleftrightarrow 2\pi\delta(w-w_o)$$
So the fourier transform is $$S(jw) = (2\pi/T)\sum_{n=-\infty}^{\infty} \delta(w-nw_o)$$
This should help.
