Prove: $(a^{-1})^{-1}=a$ Over A Field let $0\neq a\in \mathbb{F}$ 
Prove: $$(a^{-1})^{-1}=a$$
I have tried to start with the definition of an inverse, For all  $0\neq a\in \mathbb{F}$  we have:
$$a\cdot a^{-1}=a^{-1}\cdot a=1$$ But could not find a way to continue (tried multiplying by  $(a^{-1})^{-1}$
 A: $$(a^{-1})^{-1}(a^{-1})=1\implies (a^{-1})^{-1}(a^{-1})a=(a^{-1})^{-1}=a$$
A: By definition of the inverse the identity $a \cdot a^{-1}=a^{-1} \cdot a=1$ implies that $a$ is the inverse of $a^{-1}$.
A: \begin{align}
a&=1\cdot a\tag{definition of 1}\\
&=((a^{-1})^{-1}\cdot (a^{-1}))\cdot a\tag{definition of inverse}\\
&=(a^{-1})^{-1}\cdot ((a^{-1})\cdot a)\tag{associativity}\\
&=(a^{-1})^{-1}\cdot 1\tag{definition of inverse}\\
&=(a^{-1})^{-1}\tag{definition of 1}
\end{align}
A: There are several correct answers here. I will try another, using more words, in hopes of making things clearer.
By definition, the inverse of an element $a$ is the thing you multiply $a$ by to get $1$. Suppose the inverse of $a$ is $b$. That means $ab = 1$. So $a$ is what you multiply $b$ by to get $1$ so $a$ is the inverse of $b$. Inverses come in pairs, e.g. $2$ and $1/2$ in the rational numbers.
Now if you choose to name the inverse of $x$ with the notation $x^{-1}$ then it follows from the previous statement that $(a^{-1})^{-1}= b^{-1} = a$.
Since you are working in a field, you do not need to worry about whether 
an inverse exists or is unique or works with multiplication in either order. Those complications come up in other algebraic structures.
A: Since a has an inverse then we have 
          $$a•a^{-1}=a^{-1}•a=1$$
Then we  just replace a by b to get 
        $$ b•a^{-1}=a^{-1}•b=1$$
Which gives us that b is actually the inverse of $a^{-1}$
Thus: $$ b={a^{-1}}^{-1}$$
Hence done.
Hope it helps
