I have been trying to work out this strling number problem using some other recurrence which I had worked on before $P_{n,k}=1 + P_{n-1,k-1}+ P_{n-1,k}$. One solution was... $$P_{n, k}=\sum_{j=1}^{k}{n \choose j} -->(1)$$
Now I'm trying to apply this solution to find a solution to Stirling number of the second kind recurrence( which I already found in Wiki) $$S_{n,k}=S_{n-1,k-1} + kS_{n-1,k}->(2)$$ like this. $$P_{n,k}=1 + P_{n-1,k-1}+ P_{n-1,k}$$ $$=>kP_{n,k}=k + kP_{n-1,k-1}+ kP_{n-1,k}$$ $$=>kP_{n,k} - k= P_{n-1,k-1}+ (k-1)P_{n-1,k-1} + kP_{n-1,k}$$ $$=>kP_{n,k} - k - (k-1)P_{n-1,k-1}= P_{n-1,k-1} + kP_{n-1,k}$$
Now, the right hand side is exactly the right hand side of the Stirling recurrence from (1). But, substituting from (1) doesn't match the solution for (2) which is $$\frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}{k \choose j}{j^n}$$ $$\ne{k}{\sum_{j=1}^{k}{n \choose j}} - (k-1)\sum_{j=1}^{k-1}{{n-1} \choose j}-k$$
So what's wrong? Thanks!