Calculating the multiplicative inverse of a number in $GF(2^n)$ where $n > 8$
First, let us take AES as an example.

We have $\operatorname{GF}(2^8)$ where each bit represent the coefficient of a polynomial in $\operatorname{GF}(2)[X]$. The modulo reduction is done with $$p(x) = x^8 + x^4 + x^3 + x+ 1$$

Thus you can generate all the elements from a $1$ to $255$ and this forms a cyclic group : $&lt;g^1, g^{\ldots}, g^{255}&gt;$

Notice that:
$g^{2^8-1} = g \iff g^{255} = g$
or:
$g \cdot g^{2^8-2} = 1 \iff g \cdot g^{254} = 1$
thus $g^{254}$ is the inverse of $g$ or in other words:

Computing the $254^{th}$ power of an element give its inverse (more here)

This exponentiation can be quickly done using square-and-multiply.




  Thus for a field of $2^n$ with $p$ being an irreducible polynomial of degree $n$, for $g$ an element of $\operatorname{GF}(2)[X]$: $$g ^{-1} = g^{2^n-2} \bmod p$$


Also the use of tables is ill-advised do to the constant timeness. A bit-sliced implementation of this operation will be preferable.