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I'm currently stuck with the following problem:

Find all the roots of the equation $$1-\frac{x}{1}+ \frac{x(x-1)}{2!}-...+(-1)^n \frac{x(x-1)...(x-n+1)}{n!}=0$$

I can sort of see that the roots would be $1,2,3,...,n$ but cannot find a way to actually calculate them or show that they indeed are the roots. It'd be great if anyone can give me any suggestions. Hints would be more welcome than full answers. Thank you.

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    $\begingroup$ Prove by induction that it is $(-1)^n(x-1)(x-2)\dots(x-n)/n!$ $\endgroup$ – user8268 Apr 2 '16 at 10:04
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    $\begingroup$ The roots are $1, \dots , n$. $\endgroup$ – Crostul Apr 2 '16 at 10:05
  • $\begingroup$ @Crostul Yeah. I've mentioned that in the question. $\endgroup$ – SinTan1729 Apr 2 '16 at 10:06
  • $\begingroup$ @uset8268 Okay. I'm gonna give it a try. $\endgroup$ – SinTan1729 Apr 2 '16 at 10:07
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Call $f(x)$ your polynomial. For all $a \in \{ 1, \dots , n\}$ you have $$f(a) = \sum_{k=0}^a \binom{a}{k} (-1)^k + 0 \times \mbox{something} = \sum_{k=0}^a \binom{a}{k} (-1)^k1^{a-k} = (1-1)^a=0^a=0$$ hence, $1, \dots , n$ are roots of $f(x)$.

Since $f$ has at most $n$ roots, we found them all.

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  • $\begingroup$ Oh yeah! How could I miss that. It's a silly one. Nice solution though. $\endgroup$ – SinTan1729 Apr 2 '16 at 10:10

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