Why does the Fourier Transform of the impulse look so different from the Fourier Transform of the impulse train?
This answer shows that $\sum\limits_{n=-\infty}^{\infty}
  e^{j  n \Omega T}$ is actually the Fourier Series of $\frac{2\pi}{T} \sum_{k=-\infty}^{\infty} \delta(\Omega - k\tfrac{2\pi}{T})$.

\begin{equation}
 \delta(t-nT) \frac{2\pi}{T} e^{j  n \Omega T}
\end{equation}
which means
\begin{equation}
 \sum\limits_{n = -\infty}^{\infty} \delta(t-nT) \longleftrightarrow \sum\limits_{n = -\infty}^{\infty} e^{j  n \Omega T}
\end{equation}
Now consider the exponential Fourier series of $\frac{2\pi}{ T} \sum_{k=-\infty}^{\infty} \delta(\Omega - k\Omega_s)$, because this function is periodic, so we can say
\begin{equation}
 \frac{2\pi}{ T} \sum_{k=-\infty}^{\infty} \delta(\Omega - k\Omega_s)
 =
 \sum\limits_{n=-\infty}^{\infty}
 c_n e^{j  n \Omega T}
\end{equation}
where 
\begin{equation}
 c_n = \frac{1}{\Omega_s}\int\limits_{-\frac{\Omega_s}{2}}^{\frac{\Omega_s}{2}}
 \frac{2\pi}{ T} \sum_{k=-\infty}^{\infty} \delta(\Omega - k\Omega_s)
 e^{j  n \Omega T}
 d (\Omega T)
\end{equation}
and $\Omega_s = \frac{2\pi}{T}$ by definition.

Take $\alpha = \Omega T$,
\begin{equation}
 c_n = \frac{1}{\Omega_s}
 \frac{2\pi}{ T} \sum_{k=-\infty}^{\infty} 
 \int\limits_{-\frac{T\Omega_s}{2}}^{\frac{T\Omega_s}{2}}\delta(\frac{\alpha
}{T} - k\Omega_s)
 e^{-j  n \alpha}
 d \alpha
\end{equation}
The term $\frac{1}{\Omega_s}
 \frac{2\pi}{T}  = 1$ and take now
\begin{equation}
 \beta = \frac{\alpha}{T}
\end{equation}
\begin{equation}
 c_n = 
  \sum_{k=-\infty}^{\infty} 
 \int\limits_{-\frac{\Omega_s}{2}}^{\frac{\Omega_s}{2}}\delta(\beta - k\Omega_s)
 e^{-j  n \beta T}
 d \beta
 \tag{1}
\end{equation}
We know that 
\begin{equation}
 \int_{a}^b \delta(x - x_0) f(x) \ dx
 =
 f(x_0)
 \quad
 \text{if } a &lt; x_0 &lt; b
\end{equation}
and zero elsewhere. So the integral in $(1)$ contains one value within $[-\frac{\Omega_s}{2},\frac{\Omega_s}{2}]$ i.e. for $k = 0$, so 
\begin{equation}
 c_n = 
  \sum_{k=-\infty}^{\infty} 
 \int\limits_{-\frac{\Omega_s}{2}}^{\frac{\Omega_s}{2}}\delta(\beta - k\Omega_s)
 e^{-j  n \beta T}
 d \beta
=
e^{-j  n (0) T}
=
1
\end{equation}
So $c_n = 1$ for all $n$. We have just proved that 
\begin{equation}
 \frac{2\pi}{ T} \sum_{k=-\infty}^{\infty} \delta(\Omega - k\Omega_s)
 =
 \sum\limits_{n=-\infty}^{\infty}
  e^{j  n \Omega T}
\end{equation}
So whether you say $\sum\limits_{n=-\infty}^{\infty}
  e^{j  n \Omega T}$ or $\frac{2\pi}{ T} \sum_{k=-\infty}^{\infty} \delta(\Omega - k\Omega_s)$, it is the same thing.