On the Wikipedia I found the following formula:
$ x^m = \sum\limits_{k=1}^{m+1} \left \{ \begin{array}{cols} m+1 \\ k \end{array}\right \} (x-1)^{\underline{k-1}} \,\,\,\,\,\, (1) $
See: https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind#Generating_functions
Now on another forum the question was asked if the formula below is correct:
$ x^m = \sum\limits_{i=0}^m \left ( \begin{array}{cols} x-1 \\ i \end{array}\right ) \cdot \sum\limits_{j=0}^i \left ( \begin{array}{cols} i \\ j \end{array}\right ) (-1)^j (i+1-j)^m \,\,\,\,\,\, (2) $
After much trouble I may have found a derivation of formula (2) from formula (1). But I'm only an amateur, and I would like to hear from some experts whether or not my derivation is correct.
This is my derivation:
$ x^m = \sum\limits_{k=1}^{m+1} \left \{ \begin{array}{cols} m+1 \\ k \end{array}\right \} (x-1)^{\underline{k-1}} $
[Change of dummy variable]
$ x^m = \sum\limits_{i=0}^m (x-1)^{\underline{i}} \cdot \left \{ \begin{array}{cols} m+1 \\ i+1 \end{array}\right \} $
[Definition of Stirling numbers of the second kind.]
$ x^m = \sum\limits_{i=0}^m (x-1)^{\underline{i}} \cdot \frac{1}{(i+1)!} \sum\limits_{j=0}^{i+1} (-1)^j \left ( \begin{array}{cols} i+1 \\ j \end{array}\right ) (i+1-j)^{m+1} $
[Term for j=i+1 is zero.]
$ x^m = \sum\limits_{i=0}^m (x-1)^{\underline{i}} \cdot \frac{1}{(i+1)!} \sum\limits_{j=0}^i (-1)^j \left ( \begin{array}{cols} i+1 \\ j \end{array}\right ) (i+1-j)^{m+1} $
[(i+1)! = i!.(i+1)]
$ x^m = \sum\limits_{i=0}^m \frac{(x-1)^{\underline{i}}}{ i! } \cdot \frac{1}{i+1} \sum\limits_{j=0}^i (-1)^j \left ( \begin{array}{cols} i+1 \\ j \end{array}\right ) (i+1-j)^{m+1} $
[Identity for the binomial coefficient.]
$ x^m = \sum\limits_{i=0}^m \left ( \begin{array}{cols} x-1 \\ i \end{array}\right ) \cdot \frac{1}{i+1} \sum\limits_{j=0}^i (-1)^j \left ( \begin{array}{cols} i+1 \\ j \end{array}\right ) (i+1-j)^{m+1} $
[(i+1-j)^(m+1) = (i+1-j).(i+1-j)^m.]
$ x^m = \sum\limits_{i=0}^m \left ( \begin{array}{cols} x-1 \\ i \end{array}\right ) \cdot \sum\limits_{j=0}^i \left ( \begin{array}{cols} i+1 \\ j \end{array}\right ) \frac{i+1-j}{i+1} \, (-1)^j (i+1-j)^m $
[Identity for the binomial coefficient.]
$ x^m = \sum\limits_{i=0}^m \left ( \begin{array}{cols} x-1 \\ i \end{array}\right ) \cdot \sum\limits_{j=0}^i \left ( \begin{array}{cols} i \\ j \end{array}\right ) (-1)^j (i+1-j)^m $