Complexity of Discrete Logarithms in binary fields without special exponents or special structure There has been recent work, due to Joux, Gologlu, Zumbragel and others which has developed efficient algorithms for discrete logarithms in small (and specifically binary) characteristics, where the exponents have some special form. See the discussion in the questions 

how-robust-is-discrete-logarithm-in-GF(2^n) 

and 

do-recent-announcements-about-solving-the-dlp-in-gf26120-apply-to-schemes?noredirect=1&amp;lq=1

My understanding is that discrete logarithms in $\operatorname{GF}(2^n)$ where $n$ is large and has no special for is still relatively robust to attacks.

Here is the question, say I pick $n$ large enough such that the multiplicative group $\operatorname{GF}(2^n)^{\ast}$ has no small subgroups. So either $2^n-1$ is prime, or $n$ is so large and well chosen that the largest subgroup is large enough.

Would the best algorithm in this case still be baby-step-giant-step for the discrete logarithm? Or something else of still exponential (in $n$) time and memory complexity?
Edit: My hunch that the best complexity is exponential was incorrect. It is actually quasi-polynomial (thanks to Samuel Neves for his comment).

Recall that  Coppersmith's algorithm has complexity
$$
L_{2^n}(1/3,c)
$$
for a small constant $c,$ while even for $n$ prime, which is the hardest case,  discrete logarithms over $GF(2^n)$ using Barbulescu et. al.'s method   have quasi polynomial complexity
$$
O(2^{c(\log n)^2})\asymp c' n^{c \log n}.
$$

Note that the authors state "The crossing point between the (best knoen before) L(1/4) algorithm and our quasi-polynomial one is not determined yet".