Creation operator acting on a coherent state. Occupation number operator For a coherent state
$$|\alpha\rangle=e^{-\frac{|\alpha|^{2}}{2}}\sum_{n=0}^{\infty}\frac{\alpha^{n}(a^{\dagger})^n}{n!}|0\rangle$$
I want to find a simplified expression for $a^{\dagger}|\alpha\rangle.$ I can only get this $$\begin{align}
a^{\dagger}|\alpha\rangle=e^{-\frac{|\alpha|^{2}}{2}}\sum_{n=0}^{\infty}\frac{\alpha^{n}(a^{\dagger})^{n+1}}{n!}|0\rangle=e^{-\frac{|\alpha|^{2}}{2}}\sum_{n=0}^{\infty}\frac{\alpha^{n}}{\sqrt{n!}}\sqrt{n+1}|n+1\rangle
\end{align}$$
or $$a^{\dagger}|\alpha\rangle=e^{-\frac{|\alpha|^{2}}{2}}a^{\dagger}e^{\alpha a^{\dagger}}|0\rangle.$$ Is it possible to get something more "beautiful" and "useful"?(I apologize for the unscientific lexicon.)
Ultimately, I want to find a simplified expression for $N|\alpha\rangle=a^{\dagger}a|\alpha\rangle,$  but I don't know such an expression for $a^{\dagger}|\alpha\rangle.$
 A: There's no easy expression for $a^\dagger\vert\alpha\rangle$ but you are interested in $\hat N\vert \alpha\rangle$ the easy way is
\begin{align}
\hat N\vert\alpha\rangle &= \hat N e^{-\vert\alpha\vert^2/2}
\sum_n \frac{\alpha^n}{\sqrt{n!}}\hat N\vert n\rangle\, ,\\
&= \hat N e^{-\vert\alpha\vert^2/2}
\sum_n \frac{\alpha^n}{\sqrt{n!}}n\vert n\rangle\, . \tag{1}
\end{align}
What is simple and useful is 
$$
\langle \alpha\vert a^\dagger =\alpha^*\langle \alpha\vert \tag{2}
$$
obtained by taking the transpose conjugate of $a\vert\alpha\rangle=\alpha\vert\alpha\rangle$.  The calculation of $\langle N\rangle$ then easily follows.
A: The following expression can sometimes be useful:
$$ a^\dagger |\alpha\rangle = \left( \partial_\alpha + \frac{\alpha^\ast}{2} \right) |\alpha\rangle . $$
To prove this, just calculate
$$ \partial_\alpha |\alpha\rangle = \partial_\alpha \left( \mathrm e^{-|\alpha|^2 / 2} \sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}} |n\rangle \right) $$
using the product rule and $\partial_\alpha |\alpha|^2 = \alpha^\ast$.
