Using Carlitz's exponential formula to prove an identity This is a question on the homework for my finite fields class. The beginning of the assignment defines the following notation:
For $i\geq1 $, define the following elements of $A=\mathbb{F}_{q}[T]$ : 
$[i]=T^{q^{i}}-T$, 
$L_{i}=[i][i-1]\cdots[1]$, 
$D_{i}=[i][i-1]^{q}\cdots[1]^{q^{i-1}}$ 
and put $D_{0}=L_{0}=1$. 
Let $n\geq1$.
Put $A(n):=\{a \in A\mid \text{deg} (a)< n \} $
and $e_{n}(x):=\prod_{a\in A(n)}\left(x+a\right)$.
I am trying to prove: $\sum\frac{1}{b}=\frac{(-1)^{n}}{L_{n}}$ where the LHS ranges over all monic $b$ in $A$ of degree equal to $n$.
Earlier on the assignment, we showed $e_{n}(x)+D_{n}=\prod(x+b)$ where $b$ is monic with degree $n$.
I thought I would be able to take the Carlitz analog of the logarithm $\log_{c}(x)$ and then take the formal derivative and evaluate it at zero. But I've realized it is not true that $\log_{c}(xy)=\log_{c}(x)+\log_{c}(y)$. I've been stuck on this problem for some time, and can't think of any other way to approach it. I would appreciate a nudge in the right direction. 
 A: From $e_n(x)+D_n=\prod(x+b)$
it follows that $\prod b=D_n$ and the $x^1$ coefficient of $e_n(x)$
is $D_n\sum 1/b$. But the $x^1$ coefficient is equal to $\prod'a$
where the product is over all nonzero polynomials $a$ of degree $ < n$.
All you have to do is to prove this product equals $(-1)^nD_n/L_n$.
The leading coefficient should be easy enough. For the rest you need
to do a census of the degrees of the irreducible factors of polynomials
of degree $ < n$. Possibly some earlier results in your course might
cover this?
A: In class today we asked the professor about this problem and he told us to look up logarithmic derivatives. This approach leads to the following solution:
Using the previous homework problem we know that: $e_{n}(x)+D_{n}=\prod(x+a)$
Consider $\frac{f'}{f}$ where we are formally differentiating with respect to x. This transforms the LHS to $\frac{e\prime_{n}(x)}{e_{n}(x)+D_{n}}$ where 
$ e\prime_{n}(x)=\frac{d}{dx}\sum_{i=0}^{n-1}x^{q^{i}}\frac{D_{n}}{D_{i}L_{n-i}}=(-1)^{n}\frac{D_{n}}{L_{n}}+\sum_{i=1}^{n-1}(-1)^{n-i}q^{i}\frac{D_{n}}{D_{i}L_{n-i}^{q}}x^{q^{i-1}}$. 
At $x=0$, $e\prime_{n}(x)=(-1)^{n}\frac{D_{n}}{L_{n}}$ and so the LHS becomes $\frac{(-1)^{n}}{L_{n}}$. 
Similarly, the RHS gets transformed to $\sum\frac{1}{x+b}$ since $f'=\sum_{b}\frac{\prod(x+a)}{x+b}$ and $f=\prod(x+a)$. At $x=0$ this becomes $\sum\frac{1}{b}$ which is what we wanted.
