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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.

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    They are geometric sums. Googling a bit, or searching in this web, you will find a lot about geometric sums. – 2017-02-28
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    The sum is $1$ if for example $a=0$, or $n=1\,$. What are the conditions you are assuming? – 2017-02-28
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    I am working on proving that if $a^n-1$ is prime then a=2 and n is prime. Basically, $a^n-1=(a-1)(a^{n-1}+a^{n-2}+\dots +a +1$. since $a^n-1$ is prime, one of the factors has to be 1, so I am trying to show that $(a-1)=1 \rightarrow a=2$. I need that $(a^{n-1}+a^{n-2}+\dots +a +1$. since $a^n-1$ ins't one I think. – 2017-02-28
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    @Math4Life Then you have already that $a \ge 1$ and $n \ge 2\,$ so the sum is always $\ge 2\,$. – 2017-02-28
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    What makes you say that I already have those conditions? I am having trouble seeing that... thank you for the help – 2017-02-28
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    The statement you say you are trying to prove only makes sense for integer $a \ge 1$ (in fact $a \ge 2$ since $1^n-1=0$) and $n \ge 2\,$. But that's yours to spell out in the question, rather than leave to others to guess. – 2017-02-28
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    If $a = -1$ and $n = \text{odd}$, then you have equality, not inequality. – 2017-02-28

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$\sum^{n-1}_{k=0} a^k = a^{n-1}+a^{n-2}+\dots+a+1= \frac{1-a^n}{1-a}$. 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.

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    Or $a = -1$ and $n$ is odd. – 2017-02-28