$F = GF(2^6)$ modulo the primitive polynomial $h(x) = 1 + x^2 + x^3 + x^5 + x^6$ and $\alpha$ is the class of $x$: $GF(2^6) = \{0,1,\alpha, \alpha^2...\alpha^{62}\}$
How do I manually compute $\alpha^{53}$ (or any other large power of $\alpha$)? It is straightforward to compute e.g. $\alpha^{13}$ - I can just divide $x^{13}$ by $h(x)$ and take the reminder. I came across such a problem in a problem set and I suspect there should be an efficient technique to compute the remainder of $x^{53}$ divided by $h(x)$ by hand. Perhaps I should take into account that $53 = -11 (mod 64)$?
Here’s one general way, which may not be the most efficient in your particular case: you can treat it as a matter of repeated squaring and multiplication. Note that $53=32+16+4+1$, $110101$ in Binary notation. So $x^{53}=((((x^2\cdot x)^2)^2\cdot x)^2)^2\cdot x$, just eight multiplications. This method is good in any group at all, not just $\mathbb F_{64}^*$.