Proving by induction that $(\sum_{j=1}^{i-1}x_{n+j})^i=0$ I am working with the polynomial ring $R=\dfrac{\mathbb Z_2[x_1,\cdots,x_{2n}]}I$ where $$I=\left\langle x_i\,x_{n+i}\,,\,x_1+x_{n+1}\,,\,x_i+\sum_{j=1}^{i}x_{n+j}\text{ for } 2\leq i\leq n\right\rangle$$
For convenience we shall denote the class of $x_i$ as $x_i$ itself. 

Question : Prove that $x_{n+i}^{i+1}=0$ in $R$ for all $1\leq i\leq n$

Attempt : (by induction on $i$)
Clearly for $i=1$ we get $x_{n+1}^2=0$ as follows - $$x_{n+1}(x_1+x_{n+1})=0\implies \underbrace{x_1\,x_{n+1}}_{=\,0}+x_{n+1}^2=0\implies x_{n+1}^2=0$$
Let us assume that $x_{n+j}^{j+1}=0$ for all $1\leq j\leq i-1$ and we have to prove that $x_{n+1}^{i+1}=0$.
Now $x_{n+i}^i\left(x_i+\sum_{j=1}^{i}x_{n+j}\right)=0\implies x_{n+i}^{i+1}=\sum_{j=1}^{i-1}x_{n+j}x_{n+i}^i$
We can keep substituting for powers of $x_{n+i}$ and end up with $$x_{n+i}^{i+1}=\left(\sum_{j=1}^{i-1}x_{n+j}\right)^ix_{n+i}$$
We will be done if we can show that $$\left(\sum_{j=1}^{i-1}x_{n+j}\right)^i=0$$ I tried it for small values of $i$ and it happens to be 0 however I am struggling to prove it in general. So I tried it by induction on $i$ again but I am having trouble at the induction step -
Assuming $\left(\sum_{j=1}^{i-2}x_{n+j}\right)^{i-1}=0$ and writing $\left(\sum_{j=1}^{i-1}x_{n+j}\right)^i=\left(\sum_{j=1}^{i-2}x_{n+j}+x_{n+i-1}\right)^i$ and then use binomial expansion. But I am getting no where. It is getting too complicated. 
Thank you.
 A: Thanks to the comments by user26857, I have got the answer to this question- 
Let us restate the problem as follows - $$a_1^2=0\ ,\ a_2(a_1+a_2)=0\ ,\cdots,\ a_n(a_1+\cdots+a_n)=0\ \implies\ (a_1+\cdots+a_t)^{t+1}=0\quad\text{for } 1\leq t\leq n$$
Here addition and multiplication is mod 2.

We will show by induction that every monomial in $a_1,\cdots,a_t$ of degree $t+1$ is zero.

When $t=1$ we have $a_1^2=0$. 
Suppose every monomial in $a_1,\cdots,a_t$ of degree $t+1$ is zero for $t<k$ , we have to show that every monomial in $a_1,\cdots,a_k$ of degree $k+1$ is zero.
Let $a_1^{i_1}\,a_2^{i_2}\cdots a_k^{i_k}$ be a monomial such that $i_1+\cdots+i_k=k+1$.
Case 1 - When $i_k=0$ we have a monomial in $a_1,\cdots,a_{k-1}$ of degree $k+1$ which is zero since we already have by the induction hypothesis that such a monomial of degree $k$ is zero.
Case 2 - $i_k=1$ then removing the term $a_k$ we obtain a monomial in $a_1,\cdots,a_{k-1}$ of degree $k$ which is zero by the induction hypothesis.
Case 3 - $2\le i_k\le k+1$. We have $i_1+\cdots+i_{k-1}=k+1-i_k$. Now $$a_k(a_1+\cdots+a_k)=0\implies a_k^{i_k-1}(a_i+\cdots+a_k)=0\implies a_k^{i_k}=(a_1+\cdots+a_{k-1})^{i_k-1}a_k$$
Thus we have 
\begin{equation}
a_1^{i_1}\,a_2^{i_2}\cdots a_k^{i_k}=a_1^{i_1}\,a_2^{i_2}\cdots a_{k-1}^{i_{k-1}}(a_1+\cdots+a_{k-1})^{i_k-1}a_k \tag{1}
\end{equation}
Consider the term $(a_1+\cdots+a_{k-1})^{i_k-1}$. This is a sum of monomials $a_1^{j_1}\,a_2^{j_2}\cdots a_{k-1}^{j_{k-1}}$ where $j_1+\cdots+j_{k-1}=i_k-1$. So (1) is a sum of monomials of the form $$a_1^{i_1+j_1}\,a_2^{i_2+j_2}\cdots a_{k-1}^{i_{k-1}+j_{k-1}}a_k$$ 
Now $a_1^{i_1+j_1}\,a_2^{i_2+j_2}\cdots a_{k-1}^{i_{k-1}+j_{k-1}}$ is a monomial in $a_i,\cdots,a_{k-1}$ of degree $i_1+\cdots+i_{k-1}+j_1+\cdots j_{k-1}=k+1-i_k+i_k-1=k$ and is zero by the induction hypothesis.
Hence the result. $\square$
