For an $f$ infinitely differentiable on $(0,\infty)$ and real valued, consider a finite sum $$a_n= \sum_{r=1}^{n} {n \choose r} \frac{f^{(r-1)}(1)}{(r-1)!}$$ where $f^{(r-1)}(1)=\frac{d^{(r-1)}f(x)}{dx^{(r-1)}}|_{x=1} $ and "$!$" denotes the factorial notation.
Question: Find a closed form of $a_n$ similar to the following : $$ \sum_{r=0}^{n}\dbinom{n}{r}x^r=(1+x)^n$$
My Attempt: I tried using Binomial formula but could not get the solution.
Any help will be appreciated.
To summarise the discussion in the comments,
$$a_n = \sum_{r=1}^{n}\binom{n}{r}\frac{f^{(r-1)}(1)}{(r-1)!}$$
Note that:
$$\binom{n}{r} = \frac{n}{r} \binom{n-1}{r-1}$$
Therefore,
$$a_n = n\sum_{r=1}^{n}\binom{n-1}{r-1}\frac{f^{(r-1)}(1)}{r!}$$
$$\implies n!a_n = n\sum_{r=1}^{n}\binom{n-1}{r-1}\bigg(\frac{n!}{r!}\bigg)f^{(r-1)}(1)$$
$$= n\sum_{r=1}^{n}\binom{n-1}{r-1}\bigg(\frac{d^{(n-r)}}{{dx}^{(n-r)}}x^n\bigg)f^{(r-1)}(x)\Bigg|_{x=1}$$
$$= n\sum_{r=0}^{n-1}\binom{n-1}{r}\bigg(\frac{d^{(n-1-r)}}{{dx}^{(n-1-r)}}x^n\bigg)f^{(r)}(x)\Bigg|_{x=1}$$
$$\implies n!a_n =n\frac{d^{(n-1)}}{{dx}^{(n-1)}}\bigg(x^nf(x)\bigg)\Bigg|_{x=1}$$ by Leibniz rule. Hence:
$$a_n =\frac{1}{(n-1)!}\frac{d^{(n-1)}}{{dx}^{(n-1)}}\bigg(x^nf(x)\bigg)\Bigg|_{x=1}$$