For the Stirling numbers of the first kind, show that \begin{align*} (x)^{(n)} =\sum_{k=0}^n s'(n,k)x^k. \end{align*} For this proof we can proceed with induction, by proving the base cases first $n=0$ and $n=1$: \begin{eqnarray*} (x)^{(0)}=s'(0,0)x^0 = 1 \\ (x)^{(1)}=s'(1,1)x^1 = (1)(x)= x \end{eqnarray*}
The statement holds for the base case $n=1$. Now we must prove for every other integer, let $n=m-1$, \begin{align*} (x)^{(m-1)} =\sum_{k=0}^{m-1} s'(m-1,k)x^k \end{align*} To obtain equation for $n=m$, we use $m-1$ to find $m$ \begin{equation*} (x)^{(m)}=(x)^{(m-1)}(x+m-1) = x\cdot x^{(m-1)}+(m-1)x^{(m-1)}\end{equation*} \begin{equation*} =x\sum_{k\geq 0} s'(m-1,k)x^k + (m-1)\sum_{k\geq 0} s'(m-1,k) x^k \end{equation*} \begin{equation*} =\sum_{k\geq 0} s'(m-1,k)x^{k+1} + (m-1)\sum_{k\geq 0} s'(m-1,k) x^k \end{equation*} \begin{equation*} =\sum_{k\geq 1} s'(m-1,k)x^{k} + (m-1)\sum_{k\geq 0} s'(m-1,k) x^k \end{equation*} \begin{equation*} =\sum_{k\geq 0}[s'(m-1,k-1)+(m-1)s'(m-1,k)]x^k \end{equation*} \begin{equation*} (x)^{m} =\sum_{k\geq 0} s'(m,k)x^k \end{equation*}
this was my proof but it is not clear, i feel i did my induction wrong, any ideas of where i can improve my proof.