We're given \(\displaystyle S_n=\sum_{k=1}^{n}\frac{n!}{(k-1)!(n-k)!}\).
By multiplying the variable $k$ on top and bottom of the fraction, we get
\(\displaystyle \small S_n=\sum_{k=1}^{n}\frac{n!}{(k-1)!(n-k)!}=\sum_{k=1}^{n}\frac{k(n!)}{k(k-1)!(n-k)!}=\sum_{k=1}^{n}\frac{k(n!)}{(k)!(n-k)!}=\sum_{k=1}^{n} k {n\choose k}=\sum_{k=0}^{n} k {n\choose k}-0{n\choose k}=\sum_{k=0}^{n} k {n\choose k}\)
Since \(\displaystyle {n\choose k}={n\choose n-k}\)
We see that there is another way to rewrite $S_n$, i.e.