You can evaluate the sum of that series explicitly. Noting that $(a^\dagger)^n|0\rangle = \sqrt{n!}|n\rangle$, we have
$$e^{ a^\dagger}|0\rangle= \sum_{n=0}^\infty \frac{(a^\dagger)^n}{n!} |0\rangle = \sum_{n=1}^\infty \frac{1}{\sqrt{n!}} |n\rangle$$
which is a state with norm $\sum_n \frac{1}{n!} = e^1$.
That being said, I get the impression that you're asking how to show that $e^{a^\dagger}|\psi\rangle$ converges for an arbitrary state $|\psi\rangle$. This you cannot do, because generically it doesn't converge. $a^\dagger$ is not a bounded operator, which means that there are some $|\psi\rangle$ in your Hilbert space$^\ddagger$ on which $a^\dagger$ cannot act; the same is true for $e^{a^\dagger}$.
More explicitly, note that $|n\rangle = \frac{(a^\dagger)^n}{\sqrt{n!}} |0\rangle$, so $(a^\dagger)^m|n\rangle = \sqrt{\frac{(n+m)!}{n!}}|n+m\rangle$. As a result,
$$e^{a^\dagger} |n\rangle = \sum_{m=0}^\infty \frac{(a^\dagger)^m}{m!} |n\rangle = \sum_{m=0}^\infty \frac{\sqrt{(n+m)!}}{m!}|n+m\rangle = \sum_{\ell=n}^\infty \frac{\sqrt{\ell !}}{(\ell-n)!} |\ell\rangle$$
which is a state with norm $\sum_{\ell=n}^\infty \frac{\ell!}{[(\ell-n)!]^2} = n! _1F_1(n+1;1;1)$ via WolframAlpha. Though well-defined for any particular $|n\rangle$, this increases rapidly - faster than $n!$. As a result, if $|\psi\rangle = \sum_n c_n |n\rangle$ such that $\sum_n \bigg(n!_1 F_1(n+1;1;1)\bigg)^2|c_n|^2$ does not converge, then $|\psi\rangle$ is outside the domain of $e^{a^\dagger}$.
$^\ddagger$For example, $|\psi\rangle = \sum_{n=0}^\infty \frac{1}{n+1}|n\rangle$. This state is square-normalizable (it has norm $\frac{\pi}{\sqrt 6}$) but if you naively act with $a^\dagger$, you obtain $\sum_{n=1}^\infty \frac{1}{\sqrt{n}}|n\rangle$ whose norm is $\sum_{n=1}^\infty \frac{1}{n} \rightarrow \infty$; as a result, $a^\dagger |\psi\rangle$ is not in the Hilbert space, so $|\psi\rangle$ is not in the domain of $a^\dagger$.
