Better way to show that $a^{n-1}+a^{n-2}+\dots+a+1 \neq 1$ Better way to show that $a^{n-1}+a^{n-2}+\dots+a+1 \neq 1$ The way that I am currently trying is to show that $$a^{n-1}+a^{n-2}+\dots+a+1 \neq 1$$ $$a^{n-1}+a^{n-2}+\dots+a \neq 0$$ I then follow with a wishy washy argument about how if a is negative the terms will alternate in sign but not cancel each other out and how if a is positive we get a large number. Is there a more straightforward way to show this? Thanks.

$\sum^{n-1}_{k=0} a^k = a^{n-1}+a^{n-2}+\dots+a+1= \frac{1-a^n}{1-a}$. This is only 1 if $a=0$ or $n=1$. The conditions of the bigger problem show these conditions are not accurate. Thus in valid conditions, it is not equal to 1.