Is this a Taylor series? $\ln(x) +1 = \sum_{n=0}^{\infty} \frac{n+1}{n!} \cdot \frac{(\ln(x))^n}{x}$

In more steps than maybe necessary:

$$\sum_{n=0}^\infty{n+1\over n!}{(\ln x)^n\over x}={1\over x}\left(\sum_{n=1}^\infty{(\ln x)^n\over(n-1)!}+\sum_{n=0}^\infty{(\ln x)^n\over n!} \right)={1\over x}\left(\ln x\sum_{n=1}^\infty{(\ln x)^{n-1}\over(n-1)!}+e^{\ln x} \right)={1\over x}\left(\ln x\sum_{k=0}^\infty{(\ln x)^k\over k!}+x \right)={1\over x}\left((\ln x)e^{\ln x}+x \right)={1\over x}((\ln x)x+x)=\ln x+1$$