Stirling numbers of first kind proof 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.

The Stirling numbers of the first kind $s(n,k)$ satisfy the recurrence:
$$ s(n,k) = s(n-1,k-1) + (n-1) s(n-1,k) $$
The rising factorials are defined by:
$$(x)^n = x(x+1)(x+2) \cdots (x+n-1) $$
We wish to show that:
$$ (x)^n = \sum_k s(n,k) x^k $$
The induction hypothesis is that:
$$ (x)^{n-1} = \sum_k s(n-1,k) x^k $$
Using our recurrence we have:
$$ \sum_k s(n,k) x^k  = \sum_k s(n-1,k-1)x^k + (n-1) \sum_k s(n-1,k) x^k \\
= x \sum_k s(n-1,k-1)x^{k-1} + (n-1) (x)^{n-1} \\
= (x + n-1) (x)^{n-1} \\
= (x)^n $$
