Answer
$p(x+1)=\Sigma_{i=0}^{n}\frac{p^{i}({x})}{i!}$
Work Step by Step
$p(x+1)=\Sigma_{i=0}^{\infty}\frac{p^{i}({a})}{i!}(x+1-a)^{i}$
$=\Sigma_{i=0}^{n}\frac{p^{i}({x})}{i!}(x+1-x)^{i}$
$=\Sigma_{i=0}^{n}\frac{p^{i}({x})}{i!}(1)^{i}$
$=\Sigma_{i=0}^{n}\frac{p^{i}({x})}{i!}$
Hence,
$p(x+1)=\Sigma_{i=0}^{n}\frac{p^{i}({x})}{i!}$