Taylor series of $\sum_{k=0}^\infty \frac{x^k}{k!} \log(\Gamma(k+1))$

Question

Does a the following Taylor series has a closed form expression \begin{align} \sum_{k=0}^\infty \frac{x^k}{k!} \log(k!). \end{align}

Note that $\Gamma(k+1)=k!$.

Also, note that this power series converges (see for example here).

Answer 1

This is not answer, but it is a little long for a comment.

Let $$I(a)=\sum_{k=0}^{+\infty} \frac{x^k}{k!}\log \Gamma(k+a)$$ Then $$I'(a)=\sum_{k=0}^{+\infty} \frac{x^k}{k!}\psi^{(0)}(k+a)$$ And because $$\psi^{(0)}(k+a)=H_{k+a-1}-\gamma$$ We can find that \begin{align} I'(a)&=\sum_{k=0}^{+\infty} \frac{x^k}{k!}H_{k+a-1}-\sum_{k=0}^{+\infty}\gamma \frac{x^k}{k!} \\ &=\psi^{(0)}(1)e^x+\sum_{k=0}^{+\infty} \frac{x^k}{k!}H_{k+a-1} \end{align} However, I am not sure how to evaluate the last sum.