My problem is to prove that cyclic groups of order 3 or greater must have at least 2 generators.

I think I found an answer but I'm worried that it's wrong.

Here's my proof.

In a group of order $\displaystyle n$ generated by $\displaystyle a$ there exists a "last element" $\displaystyle a^{n-1}$. This element can be shown to generate every inverse element and therefore every element in the group.

For example $\displaystyle a^{n-1}a^{n-1}=a^{2n}a^{-2}=ea^{-2}=a^{-2}$

$\displaystyle a^{n-1}a^{n-1}a^{n-1}=a^{3n}a^{-3}=ea^{-3}=a^{-3}$

and so on until $\displaystyle a^{n-1}a^{n-1}\cdots(n times)\cdots a^{n-1}=a^{n^2}a^{n}=e$

Thus it is shown that in every cyclic group with $\displaystyle n>2$ the element $\displaystyle a^{n-1}$ generates the group as well as $\displaystyle a$.

Is this valid?