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**1. The problem statement, all variables and given/known data**

Show that the binomial coefficients ## \binom {-1}{n}=(-1)^n##

**2. Relevant equations**

##\binom{n}{k}=\frac{n!}{(n-k)!k!} \\

##

**3. The attempt at a solution**

##-1!=(-1)\cdot 1! \\

-1!=-1 \\

-2!=(-1)^2 \cdot 2! \\

-n!=(-1)^n \cdot n!\\

\mbox{for n=0} \\

LHS=\binom{-1}{0}=\frac{-1!}{(-1!)(0!)}=1 \\

RHS=(-1)^0=1 \\

LHS=RHS \space \mbox{When n=0 therefore it is valid for n=0} \\

\mbox{Assuming it holds for some integer k}\\

\binom{-1}{k}=(-1)^k \\

\mbox{Now trying to prove for k+1} \\

\binom{-1}{k+1}=(-1)^{k+1} \\

LHS=\binom{-1}{k+1}

=\frac{-1!}{[(-1-(k+1)]!(k+1)!}\\

=\frac{-1}{(-k-2)!(k+1)!} \\

=\frac{-1}{(-1)^{k+2}(k+2)!(k+1)!}\\

=\frac{1}{(-1)^{k+1}(k+2)!(k+1)!} ##

So I obviously did something wrong since I didn't get both sides being equal, it probably has to do with how I treated the negative factorials.