Signed Stirling Number of the First Kind of Number with Greater

Theorem
Let $n, k \in \Z_{\ge 0}$

Let $k > n$.

Let $s \left({n, k}\right)$ denote a signed Stirling number of the first kind.

Then:
 * $s \left({n, k}\right) = 0$

Proof
By definition, the signed Stirling numbers of the first kind are defined as the polynomial coefficients $s \left({n, k}\right)$ which satisfy the equation:


 * $\displaystyle x^{\underline n} = \sum_k s \left({n, k}\right) x^k$

where $x^{\underline n}$ denotes the $n$th falling factorial of $x$.

Both of the expressions on the and  are polynomials in $x$ of degree $n$.

Hence the coefficient $s \left({n, k}\right)$ of $x^k$ where $k > n$ is $0$.

Also see

 * Unsigned Stirling Number of the First Kind of Number with Greater equals Zero
 * Stirling Number of the Second Kind of Number with Greater equals Zero