# Stirling #s of 1st kind. Prove the following: $\sum_{k=0}^n S_1(n,k)x^k = x(x +1)···(x +n−1)$

$\mathbf {Theorem:}$ $$\sum_{k=0}^n S_1(n,k)x^k = x(x +1)···(x +n−1)$$

I want to prove the above theorem, and I know that I should use the recurrence relation $$S_1(n,k)=S_1(n-1,k-1)+(n-1)S_1(n-1,k).$$ I also know that $$\sum_{k=0}^n S_1(n,k)=n! ,\quad S_1(n,1)=(n-1)! ,\quad S_1(n,n)=1,$$ and $$S_1(n,0)=S_1(0,n)=0 \quad \text{if} \quad n\neq0.$$ I just started doing this:$$\sum_{k=0}^n S_1(n,k)x^k =S_1(n,0)+S_1(n,1)x+S_1(n,2)x^2+...+S_1(n,n)x^n.$$ However I do not know how to proceed further. Can anybody help at this point?

• Multiply the last equation by $(x+n)$ and use $S_1(n+1,k)=S_1(n,k-1)+nS_1(n,k)$ ... & of course induction. – Donald Splutterwit Jul 23 '17 at 12:20