An estimate to show that Goss L functions are entire In Section 8 of Goss' Basic Structure of Function Field Arithmetic, Goss computed an estimate in section 8.8 in order to show that his L-function is entire on $S_\infty$. The main tool is binomial/multinomial theorem. In proving (2), he said this can be done by expanding by binomial theorem and use first part. I am not sure how the result can be obtained, as after expanding I cannot get something that looks like (1). May someone explain a bit more about the details in this part of the proof?
 A: From your description it sounds like you are working through Lemma 8.8.1 in Goss's book, whose statement is fairly technical but which has several applications to the entireness of Goss $L$-functions as well as to special values of Goss $L$-functions at negative integers.
Before going on to the Goss's lemma, the fine details of which some may want to skip, I'll point out that it has several nice applications: for example (Prop. 8.8.2 in Goss) if we let $W(d) \subseteq \mathbb{F}_q[\theta]$ be the $\mathbb{F}_q$-subspace of all polynomials of degree at most $d$ with constant term $0$, then taking $v_{\theta}$ to be the valuation at $\theta=0$, we must have for any non-negative integer $i$,
$$
v_\theta \left( \sum_{f \in W(d)} f(\theta)^i \right) \geq (q-1)\frac{d(d+1)}{2},
$$
which can be related to $\theta$-adic information about the Carlitz zeta function and Dirichlet type Goss $L$-functions at negative integers.  This particular estimate can be proved directly, but Goss's lemma provides a framework to prove similar estimates in much more generality.
Now your question is in the middle, but it might make sense to go through the whole thing, though skip ahead to the second half for the answer.
Lemma 8.8.1 (1) Let $J,K$ be field extensions of $\mathbb{F}_q$.  Let $W \subseteq J$ be a finite dimensional $\mathbb{F}_q$-vector space of dimension $d$, and let $\{ \mathcal{L}_1, \ldots, \mathcal{L}_t\}$ be $\mathbb{F}_q$-linear maps $\mathcal{L}_h : J \to K$.  Finally, let $x \in J$ and let $\{ i_1, \dots, i_t \}$ be non-negative integers such that
$$
\sum_{i=1}^t i_h < (q-1)d.
$$
Then
$$
\tag{i}
\sum_{w\in W} \left( \prod_{h=1}^t \mathcal{L}_h(x+w)^{i_h} \right) = 0.
$$
Proof: This is the proof in Goss.  We fix an $\mathbb{F}_q$-basis $\{ e_1, \ldots, e_d \}$ of $W$ so that
$$
\textrm{(i)} = \sum_{c_1, \ldots, c_d \in \mathbb{F}_q} \prod_{h=1}^t (\mathcal{L}_h(x) + c_1\mathcal{L_h}(e_1) + \cdots + c_d\mathcal{L_h}(e_d))^{i_h}.
$$
Using the multinomial theorem,
$$
\textrm{(i)} = \sum_{c_1, \dots, c_d \in \mathbb{F}_q} \prod_{h=1}^t \sum_{j_{h0} + \cdots + j_{hd} = i_h} \binom{i_h}{j_{h0}, \ldots, j_{hd}} \mathcal{L}_h(x)^{j_{h0}} \mathcal{L}_h(e_1)^{j_{h1}} \cdots \mathcal{L}_h(e_d)^{j_{hd}} c_1^{j_{h1}} \cdots c_d^{j_{hd}}.
$$
Passing through the sum on the $c$'s, we see that we can group the terms together so that (i) consists of summands, each of which is an element of $K$ multiplied by a sum of the form
$$
\tag{ii}
\sum_{c_1, \ldots, c_d \in \mathbb{F}_q} c_1^{j_{11} + \cdots + j_{t1}} \cdots c_d^{j_{1d} + \cdots + j_{td}}.
$$
Now the total sum of exponents in (ii) is at most $\sum_{h=1}^t i_h$, which is $<(q-1)d$.  Therefore there must be some $\ell$ so that $j_{1\ell} + \cdots + j_{t\ell} < q-1$, so at least one of the exponents in (ii) is not divisible by $q-1$, and so (ii) $= 0$.  Thus (i) $= 0$.
Lemma 8.8.1 (2) Continuing as above, we first assume that $K$ has a discrete valuation $v$ such that $v(\mathcal{L}_h(w)) > 0$ for all $h$ and for all $w \in W$. For $j > 0$ put
$$
W_j = \{ w \in W \mid \forall h,\, v(\mathcal{L}_h(w)) \geq j\}.
$$
Then for any arbitrary collection of non-negative integers $\{i_1, \ldots,i_h\}$, we have
$$
v\left( \sum_{w\in W} \prod_{h=1}^t \mathcal{L}_h(w)^{i_h} \right) \geq (q-1)D,
$$
where $D = \sum_j \mathrm{dim}_{\mathbb{F}_q} (W_{j})$.
Proof: Because $W$ is a finite dimensional $\mathbb{F}_q$-vector space, for each $h$ the image $\mathcal{L}_h(W) \subseteq K$ is a set of bounded valuation.  Thus we can pick $j_0$ so that $W_{j_0+1} = \{0\}$ but $W_{j_0} \neq \{0\}$.  In this way we have
$$
W = W_{1} \supseteq W_2 \supseteq \cdots \supseteq W_{j_0} \supsetneq \{0\}.
$$
Now let $d_j = \dim_{\mathbb{F}_q}(W_{j})$.  I believe what Goss is saying can be proved by induction is the following.
Claim: For any $j$ with $1\leq j\leq j_0$ and for any choice of non-negative exponents $\{i_1, \dots, i_t\}$,
$$
v\left(\sum_{w \in W_{j}} \prod_{h=1}^t \mathcal{L}_h(w)^{i_h} \right) \geq (q-1)d_{j} j + (q-1)\sum_{\ell > j} d_{\ell}.
$$
We first note that the estimate in the statement of the lemma is the case $j=1$ from the claim, so we are done once we prove it.  We proceed by induction (in reverse order).
In the case $j=j_0$, we see from part (1) (taking $x=0$) that
$$
v\left(\sum_{w \in W_{j_0}} \prod_{h=1}^t \mathcal{L}_h(w)^{i_h} \right) \geq (q-1)d_{j_0}j_0.
$$
Indeed if $\sum i_h < (q-1)d_{j_0}$, then the sum in the valuation is $0$ by part (1) and the valuation itself is $\infty$; otherwise if $\sum i_h \geq (q-1)d_0$, then each term in the sum has valuation at least $(q-1)d_{j_0}j_0$ by the definition of $W_{j_0}$.
Now suppose the claim is true for some $j+1$ with $2 \leq j+1 \leq j_0$.  Pick a basis $\{e_1, \ldots, e_{d_j} \}$ of $W_j$ so that $\{e_1, \ldots, e_{d_{j+1}}\}$ is a basis of $W_{j+1}$.    We can then express every $w \in W_j$ uniquely as $w=w'' + w'$, where $w' \in W_{j+1}$ and $w''$ is in the span of $\{e_{d_{j+1}+1}, \ldots, e_{d_j}\}$ (i.e. $W_j/W_{j+1}$).  Then
$$
\begin{aligned}
\sum_{w \in W_j} \prod_{h=1}^t \mathcal{L}_h(w)^{i_h} &= \sum_{w' \in W_{j+1}} \sum_{w'' \in W_j/W_{j+1}} \prod_{h=1}^t \mathcal{L}_h(w'+w'')^{i_h} \\ &= \sum_{w' \in W_{j+1}} \sum_{w'' \in W_{j}/W_{j+1}} \prod_{h=1}^t \sum_{k_h=0}^{i_h} \binom{i_h}{k_h} \mathcal{L}_h(w'')^{k_h} \mathcal{L}_h(w')^{i_h-k_h}.
\end{aligned}
$$
All of the terms in this sum contain a factor of the form $\sum_{w' \in W_{j+1}} \prod_{h=1}^t \mathcal{L}_h(w')^{i_h-k_h}$, which by the induction hypothesis has valuation
$$
\tag{iii}
v\left(\sum_{w' \in W_{j+1}} \prod_{h=1}^t \mathcal{L}_h(w')^{i_h-k_h} \right) \geq (q-1)d_{j+1} (j+1) + (q-1)\sum_{\ell > j+1} d_{\ell}.
$$
At the same time all of the terms in the sum contain a factor of the form
$$
\sum_{w'' \in W_j/W_{j+1}} \prod_{h=1}^t \mathcal{L}_h(w'')^{k_h},
$$
to which we now apply part (1) again (with $W = W_j/W_{j+1}$).  If $\sum k_h < (q-1)(d_j-d_{j+1})$ then this necessarily vanishes, so just as in the case $j=j_0$, we find
$$
v\left(\sum_{w'' \in W_j/W_{j+1}} \prod_{h=1}^t \mathcal{L}_h(w'')^{k_h}\right) \geq (q-1)(d_j - d_{j+1}) j.
$$
Combining this with (iii) we see that
$$
\begin{aligned}
v\left(\sum_{w \in W_{j}} \prod_{h=1}^t \mathcal{L}_h(w)^{i_h} \right) &\geq (q-1)\left((d_j-d_{j+1})j + d_{j+1}(j+1) + \sum_{\ell > j+1} d_{\ell}\right)\\ &=(q-1) \left( d_j j + d_{j+1} + \sum_{\ell> j+1} d_{\ell} \right),
\end{aligned}
$$
which is exactly what we wanted.
