Our material defines generator like below.
Sometimes it is easier to define the elements of the $GF(2^n)$ field using a generator. $\{0,g,g,g^2,...,g^N\}$, where $N=2^n-2$
But I cannot understand why $N=2^n-2$ not $2^n-1$.
If $n=3$, $GF(2^3)$ has 8 elements but generators are 7 including zero. Why?
You're missing $g^0$ from your counting. It's $\{0,g^0,g,g^2,...,g^N\}$ (you had $g$ twice here instead of $g^0$). If $N = 2^n-2$, then there are $2^n-1$ different powers of $g$, and a $0$ for $2^n$ elements in total.
To take the $2^3$ example, there are (if you choose $g$ wisely) $7$ distinct powers of $g$ in $GF(8)$: $$ g^0, g, g^2, g^3, g^4, g^5, g^6 $$ If we include $0$, that makes it $8$ elements in total.
