Unsigned Stirling Number of the First Kind of Number with Greater

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

Let $k > n$.

Let $\displaystyle \left[{n \atop k}\right]$ denote an unsigned Stirling number of the first kind.

Then:
 * $\displaystyle \left[{n \atop k}\right] = 0$

Proof
By definition, unsigned Stirling number of the first kind are defined as the polynomial coefficients $\displaystyle \left[{n \atop k}\right]$ which satisfy the equation:


 * $\displaystyle x^{\underline n} = \sum_k \left({-1}\right)^{n - k} \left[{n \atop k}\right] x^k$

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 $\displaystyle \left[{n \atop k}\right]$ of $x^k$ where $k > n$ is $0$.

Also see

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