Proof: By Induction
(related to Proposition: Factorial Polynomials vs. Polynomials)
Proof of the statement by induction.
"$\Rightarrow$"
 Let $\phi(x)=a_nx^{\underline{n}}+a_{n1}x^{\underline{n1}}+\ldots+a_1x^{\underline{1}}+a_0$ be a factorial polynomial of degree $n\ge 0.$
 Base case $n=0:$
 If $\phi(x)=a_0$ then set $b_0:=a_0$ and we have $\phi(x)=b_0.$
 Induction Step $n\to n+1$
 Assume, $n\ge 0$ and any factorial polynomial $$\phi(x)=a_mx^{\underline{m}}+a_{m1}x^{\underline{m1}}+\ldots+a_1x^{\underline{1}}+a_0$$ of degree $m$ equals some polynomial $$\phi(x)=b_mx^{m}+b_{m1}x^{m1}+\ldots+b_1x^{1}+b_0$$ of degree $m$ for all $m\le n.$
 By definition of the falling factorial power, we have $$x^\underline{n+1}=x(x1)\cdots(xm)=x^{n+1}+p(x),$$ where $p(x)$ is some polynomial of degree $n.$
 By induction hypothesis, $\phi(x)=a_{n+1}x^\underline{n+1}+q(x)$ with a polynomial $q(x)$ of degree $n.$
 Thus, $\phi(x)=a_{n+1}(x^{n+1}+p(x))+q(x)=a_{n+1}x^{m+1}+s(x),$ where $s(x)=a_{n+1}p(x)+q(x)$ is a polynomial of degree $n.$
 Thus, $\phi(x)$ can be written as a polynomial of degree $n+1$.
"$\Leftarrow$"
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References
Bibliography
 Miller, Kenneth S.: "An Introduction to the Calculus of Finite Differences And Difference Equations", Dover Publications, Inc, 1960