Artin-Hasse exponential and computation Let $p :=2$, and $\mathbb{F}_p[[ t ]]$ be the ring of formal power series in one variable $t$ over the finite field $\mathbb{F}_p$ with $p$ elements.  The Artin-Hasse exponential $E(t)$ is defined to be 
$$
E(t) = \textrm{exp}(- t - t^p/p - t^{p^2}/p^2 - \ldots) 
$$
It is well-known the coefficients of $E(t)$ belong to $\mathbb{Z}_{(p)} = \{ a/b : p \nmid b , a, b \in \mathbb{Z} \}$, so it has a well-defined image $\bar{E}(t) \in \mathbb{F}_p[[t]]$. 
Given $u \in t(1 + t F_2[[t ]]) $, is there any way to factorize $\bar{E}(u(t))$ in terms of integral powers of $\bar{E_n}(t) := \bar{E}(t^n)$ where $n \in \mathbb{N}$?
 A: If I understand your question, I’ve actually given some thought to just that matter. Let me just ramble — whether what I have to say is of any help will be for you to decide.
Your $u$ is, as I’m sure you know, an invertible series (in the sense of substitution), and you may have even seen the group of all such series called the “Nottingham Group”, though this is not a standard terminology. Nothing is special about the prime $2$ here, your question makes perfectly good sense, and is equally interesting over any field $\Bbb F_p$.
The multiplicative group $U=1+t\Bbb F[[t]]$ is a module over $\Bbb Z_p$, because it makes good sense to raise an element of $U$ to a power that is a $p$-adic integer. As I guess you also already know, $U$ is the direct product (not sum!) of infinitely many copies of $\Bbb Z_p$, and the A-H series allow you to write any element of $U$ uniquely as 
$$
\prod_{\gcd(p,n)=1}\bigl(E(t^n)\bigr)^{a_n}\,,
$$
where each $a_n\in\Bbb Z_p$, and where I’ve allowed myself to omit the bar over the $E$.
Now, when you write $E(u(t))$, you’re applying to $E(t)$ the automorphism of $\Bbb F_p((t))$ that’s defined by composing series (even Laurent series!) with $u$ on the right. And by asking your question, you’re asking what the effect of $u$ is on the first “basis element” $E(t)$. In the case that $p=2$, you should, by rights, also ask for the effect of $u$ on every $E(t^m)$, with $m$ running through the odd integers. So far so good. Now suppose you made a doubly infinite matrix with the exponents for $E(t)\circ u$ in the first column, the exponents for $E(t^3)\circ u$ in the second column, of $E(t^5)\circ u$ in the next, etc. I hope you see what I’m doing: mapping the Nottingham group into the group of such doubly infinite matrices. And it’s a homomorphism, a kind of infinite-dimensional representation of Nottingham. (You will need to convince yourself that the convergences work out all right.)
You’ll notice that I haven’t said a word that would help you to answer your question. And that’s because I don’t know how to answer it. If you have a specific $u$, then of course you may use a symbolic calculation program to find the “entries” $a_n$, although I’ve tried it and found that you need to calculate modulo an extremely high power of $t$ to get even a little accuracy for a few values of $n$.
I have some other ideas on this subject, and will be happy to share them with you — please feel free to e-mail me.
