Show that $(1+a_1x+\ldots+a_rx^r)^k=1+x+x^{r+1}q(x)$ 
Fixed $k\ge 1$. Show that for each $r$, you can find $a_1,\cdot\cdot\cdot,a_r\in \mathbb{F}$ such that :$$(1+a_1x+\cdot\cdot\cdot+a_rx^r)^k=1+x+x^{r+1}q(x)$$
  where $q(x)$ is a polynomial.

Any ideas?
I tried using induction as follows:  
for the base case, $r=1:$
We wish that $$(1+a_1x)^k = 1+x+x^2q(x)$$
Setting $a_1 = 1$ we get:
$$1 + x + \text{other terms} = 1 + x + x^2q(x)$$
$$\text{other terms} = x^2q(x)$$
 A: If the field $\mathbb{F}$ has characteristic $p \neq 0$, and $p \mid k$, then the requested $a_1,\dots, a_r$ do not exist: writing $k=pk'$, we have 
\begin{align*}
(1+a_1x+\cdots+a_rx^r)^k &= ((1+a_1x+\cdots+a_rx^r)^p)^{k'}\\
&= (1+a_1^px^p+\cdots+a_r^px^{rp})^{k'}
\end{align*}
which has a zero coefficient of $x$ and hence cannot be put into the form $1+x+x^2q(x)$.
So assume that $\mathbb{F}$ has characteristic 0 or that $p \nmid k$. To rephrase the problem, we are looking for a polynomial $f$ of degree $r$ with constant term 1 such that 
$$f^k \equiv 1+x \pmod{x^{r+1}}$$
As suggested, we can find such a polynomial by induction on $r$. The base case $r=0$ is trivial (just take $f=1$). For the inductive step, assume we have a polynomial $f$ of degree $r$ satisfying the required conditions. Then
$$f^k \equiv 1+x+cx^{r+1} \pmod{x^{r+2}}$$
for some $c\in\mathbb{F}$. If we set $g=f+a_{r+1}x^{r+1}$ (where $a_{r+1}$ is yet to be determined), then by the binomial theorem, and using the fact that $f$ (and hence $f^{k-1}$) has constant term 1,
\begin{align*}
g^k &= (f+a_{r+1}x^{r+1})^k \\
&\equiv f^k + ka_{r+1}f^{k-1}x^{r+1} \\
&\equiv 1+x+(c+ka_{r+1})x^{r+1} \pmod{x^{r+2}}
\end{align*}
If we set $a_{r+1}=-c/k$, then $g^k$ satisfies the required conditions, completing the proof.
