(related to Problem: Interpolating Numbers With a Polynomial)
Surprisingly, we can find a combinatorial solution to this problem using the discrete difference calculus methods. * By "Taylor's formula for the difference operator", we have for any factorial polynomial $$f(x)=\frac{\Delta^n f(0)}{n !} x^{\underline{n}}+\frac{\Delta^n f(0)}{(n-1) !} x^{\underline{n-1}}+ \ldots + \frac{\Delta^2 f(0)}{2 !} x^{\underline{2}} + \Delta f(0) x^{\underline{1}}+ f(0).$$ * Because factorial polynomials and usual polynomials equal each other, the only difference are their different coeffecients. * Since we want to find the coefficients of the of a "usual" polynomial, and not a factorial polynomial, we can, nevertheless use the above Taylor's formula as a starting point. * Writing the data in tabular form and computing the successive differences until all become $0$ yields $$\begin{array}{rrrrrrr} x&f(x)&\Delta f(x)&\Delta^2 f(x)&\Delta^3 f(x)&\Delta^4 f(x)&\Delta^7 f(x)\\ \hline 0&0&15&50&60&24&0\\ 1&15&65&110&84&24&0\\ 2&80&175&194&108&24&0\\ 3&255&369&302&132&24\\ 4&624&671&434&156\\ 5&1295&1105&590\\ 6&2400&1695\\ 7&4095 \end{array}$$ * It follows that the factorial polynomial representation of $f(x)$ is $$\begin{align}f(x)&=0+15x^{\underline 1}+\frac{50}{2!}^{\underline 2}+\frac{60}{3!}^{\underline 3}+\frac{24}{4!}^{\underline 4}\nonumber\\ &=0+15x+\frac{50}{2!}x(x-1)+\frac{60}{3!}x(x-1)(x-2)+\frac{24}{4!}x(x-1)(x-2)(x-3)\nonumber\\ &=x^4+4x^3+6x^2+4x\nonumber\\ &=(x+1)^4-1 \end{align}$$ which is the required usual representation of the polynomial.
