1. ## proof

Can somebody check my solution please?

$\displaystyle 1 - x/1! + x(x-1)/2! +..+(-1)^n x(x-1)(x-2)....(x-n+1)/n!$
$\displaystyle = (-1)^n x(x-1)(x-2)....(x-n)/n!$
Proof: by using derivation I will show that 1,2,3,,,n are simple roots of both polynomials, that implies they must be equal
Am I right? Thank you very much.

2. Originally Posted by sidi
Can somebody check my solution please?

$\displaystyle 1 - x/1! + x(x-1)/2! +..+(-1)^n x(x-1)(x-2)....(x-n+1)/n!$
$\displaystyle = (-1)^n x(x-1)(x-2)....(x-n)/n!$
Proof: by using derivation I will show that 1,2,3,,,n are simple roots of both polynomials, that implies they must be equal
Am I right? Thank you very much.
the RHS is incorrect! it should be $\displaystyle (-1)^n(x-1)(x-2) \cdots (x-n)/n!.$ of course if you can prove that 1, 2, ... , n are the roots of the LHS, then you're done, because the degrees and the leading

coefficients of both sides are equal. (what does this equation have anything to do with "advanced" algebra anyway??)

Edit: on second thought, i think it's easier to prove the identity by induction (over $\displaystyle n$ of course!).