$a$ and $1+a^{-1}$ have same degree over $F$ if $a$ is algebraic over $F$ 
Suppose that $a$ is algebraic over a field $F$. Show that $a$ and $1+a^{-1}$ have the same degree over $F$.

Not really sure where to start for this one.
I know that I have to show that $[F(a):F]=[F(1+a^{-1}):F]$, and I assume that being told that $a$ is algebraic over $F$ gives me some info, but I'm not sure how to use this info.
$a$ algebraic over $F$ means that $a$ is a zero of some non-zero polynomial with coefficients in $F$.
So there exists some minimal polynomial of the form $p(x)=c_0+c_1 x+\ldots+ c_{n-1} x^{n-1}$ for $c_i\in F$ such that $p(a)=0$. We also know that $F(a)$ is isomorphic to $F[x]/\langle p(x)\rangle$.
Guidance how to approach this problem would be appreciated. Thank you.
 A: It is obvious that $F(1+a^{-1})\subseteq F(a)$. And since
$$\frac1{(1+a^{-1})-1}=a$$
we have that $F(a)\subseteq F(1+a^{-1})$.
Thus $F(a)=F(1+a^{-1})$
They have the same index because they are the same extension.
A: How about this solution. 
It follows your idea.
Note that $F(a)=F(a^{-1})$.
Let $m(x)$ be the minimal polynomial of $a^{-1}$ over $F$.
Then $m(x-1)$ is irreducible over $F$
and $m(x-1)$ is the minimal polynomial of $1+a^{-1}$ over $F$
and $[F(1+a^{-1}):F]=\deg{m(x-1)}=\deg{m(x)}=[F(a^{-1}):F]=[F(a):F]$.
This question appears at the Gallian's Contemporary Abstract Algebra 8/e (Exercise 20.36 and 21.41).
A: Let $f(x)$ be the minimal polynomial of $a$.
Since $f(x)$ is irreducible over $F$, 
the constant term of $f(x)$ is nonzero and 
the reciprocal polynomial $f^*(x)=x^n f(1/x)$ of $f(x)$ is also irreducible over $F$ (See here) and $\deg{f^*(x)}=n$, 
where $n=\deg{f(x)}$.
Note that $f^*(a^{-1})=(a^{-1})^n f(a)=0$.
Hence, $g(x)=f^*(x)$ is the minimal polynomial of $a^{-1}$.
Furthermore, 
$g(x-1)$ is irreducible over $F$ and $g(x-1)$ is the minimal polynomial of $1+a^{-1}$.
