TITLE: Find all roots of the equation $1-\frac{x}{1}+\frac{x(x-1)}{2!}-\cdots+(-1)^n\frac{x(x-1)(x-2)...(x-n+1)}{n!}=0$
QUESTION [5 upvotes]: $ 1-\dfrac{x}{1}+\dfrac{x(x-1)}{2!}-\cdots+(-1)^n\dfrac{x(x-1)(x-2)...(x-n+1)}{n!}=0$  
I could not attempt the question at all but just rewriting it as
$P(x)=\dfrac{1}{n!}\prod\limits_{k=1}^n (k-x)$ 
But, I am not yet able to figure out how will that help. Please help.

REPLY [8 votes]: Let us first find the integer solutions to this equation ( if any ). You can rewrite the equation as $$1-\binom{x}{1}+\binom{x}{2}-\cdots+(-1)^{n}\binom{x}{n}=0$$ If $x\le n$, then the equation reads as $$1-\binom{x}{1}+\binom{x}{2}-\cdots+(-1)^x\binom{x}{x}=0\\ \Rightarrow (1-1)^x=0$$ which is true $\forall x\in \{1,2,\cdots,n\}$. But that means we have found $n$ roots of the equation which clearly is a $n$ degree polynomial and thus can have at most $n$ real roots. So, $x=1,2,\cdots,\  n$ are the roots of this equation.