Usually, the Stirling numbers of the first kind are defined as the coefficients of the rising factorial: $(*) \prod_{i=0}^{n-1}(x+i) = \sum_{i=0}^{n} S(n,i) x^i$.
With this definition, a recursive relation of $S(n,i)$ be derived, and it can be shown that is coincides with the recursive relation on the number of permutation on an n-set with i cycles and they have the same initial conditions, hence they coincide.
1) Is there any possibility to do it the other way around, i.e., define $S(n,i)$ combinatorially and then to show that $\prod_{i=0}^{n-1}(x+i) = \sum_{i=0}^{n} S(n,i) x^i$ holds for $x \in \mathbb{N}$ by some combinatorial argument, and thus it is a polynomial identity?
(For Stirling numbers of the second kind, it is possible: it can be shown that $n^k = \sum_{i=0}^{k} \binom{n}{i} i! S_2(k,i)$ combinatorially ($n^k$ counts function from $[k]$ to $[n]$).)
2) Additionally: equating coefficients in $(*)$ shows that $S(n,i)$ is the elementary symmetric polynomial on $n$ variables of degree $n-i$ evaluated on $(0,1,\cdots ,n-1)$. Is there a combinatorial interpretation of this?