Different forms of Stirling numbers of second type formula

Question

The number of partitions of a set with $m$ elements into $n$ nonempty subsets is given by the Stirling numbers of the second type, $S(m,n)$. The formula to find $S(m,n)$ is said to be:

$$ S(m,n)=\frac{1}{n!}\sum_{k=0}^n (-1)^k\binom{n}{k}(n-k)^m $$

But at some places it is given as:

$$ S(m,n)=\frac{1}{n!}\sum_{k=0}^n (-1)^{n-k}\binom{n}{k}k^m $$

How do I go to the second formula from the first one ?

Answer 1

We can go from one formula to the other by reverting the order of summation $k\rightarrow n-k$. In general we have by switching the summation order this way: \begin{align*} \sum_{k=0}^na_k =\sum_{k=0}^na_{n-k} \end{align*}

Here we obtain \begin{align*} \color{blue}{\frac{1}{n!}}&\color{blue}{\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^m}\\ &=\frac{1}{n!}\sum_{k=0}^n(-1)^{n-k}\binom{n}{n-k}(n-(n-k))^m\tag{$k\rightarrow n-k$}\\ &\color{blue}{=\frac{1}{n!}\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}k^m}\\ \end{align*}

Answer 2

Unless i'm confused, this comes directly from symmetry of binomial expansion. You can notice the symmetry by letting n=2, n=3 etc