Show that [r=0,n]Σ(-1)^r nCr=0

So I am thinking :
Since (a+b)^n=[r=0, n]Σ nCr a^n-1*b^r

a=1 and b=(-1) in r=0,n]Σ(-1)^r nCr=0 therefore (1+(-1)^n=0

Do you think this way of showing that [r=0,n]Σ(-1)^r nCr=0 would fly?

You cannot conclude that 1+(-1)^n=0
but you have proved that [r=0,n]Σ(-1)^r nCr=0