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Proving that a number has a multiplicative inverse

Is there a way to prove, without using calculator, that $a^n-1$ has a multiplicative inverse?
If I start with $a^n-1=b$ I would have to prove that $b$ is not the additive inverse of $a^n-1$ and that there is not $c$ such that $a^n-1=bc$.
But how can I prove that $b$ is not the additive inverse of $a^n-1$ without the calculator?
Is there a way to prove that there is not $c$ such that $a^n-1=bc$ without using the calculator?

A:

Let $a$ be a fixed natural number.
If $n$ is odd then we have
$$a^{n}-1 = \underbrace{a^n}_{\in\mathbb{N}}-1 = (a-1)(a^{n-1}+\cdots+a+1)$$
so
$$\gcd(a^n-1,a-1) = \gcd(a^n,a-1)=a-1.$$
If $n$ is even then
$$a^{n}-1 = (a-1) \cdot a^{n-1}+\cdots+a+1,$$
so
$$\gcd(a^n-1,a-1) = \gcd(a^n,a-1)=a-1.$$
Thus, in either case we have $\gcd(a^n-1,a-1)=a-1$.
Since $\gcd(a^n-1,a-1)=a-1$, it follows that there exists $b$ such that $b\cdot a^n-1=1$, and there exists $c$ such that $c\cdot (a^n-1)=b$.
Now, $a^n-1=b$ is equivalent to $a^n=1+b$ and therefore $a$ is a root of the polynomial $x^n-1$.

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