Is there a function that satisfies this Taylor series?

Question

I came across this today in a quiz, where I was asked to find the value of the 1st and 34th derivatives at $x=0$

Here's the series we were given:

$$f(x) = 1 + \frac{x}{2!} + \frac{x^2}{3!} + \frac{x^3}{4!} + ... + \frac{x^n}{(n+1)!}$$

I concluded that $f'(0)=\frac{1}{2}$, $f^{(34)}(0)=\frac{1}{35}$, from that, and wondered if there was a function that satisfied $f(0) = 1$, $f'(0)=\frac{1}{2}$, $f''(0)=\frac{1}{3}$, $f^{(n)}(0)=\frac{1}{n+1}$

Answer 1

Hint: Multiply the function by $x$ and add $1$. The result should look familiar.