# 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 $$a^{\dagger}|\alpha\rangle.$$

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.
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 $$\partial_\alpha |\alpha|^2 = \alpha^\ast$$.