Every nilpotent left ideal is contained in a nilpotent 2 sided ideal. Question:
Let $R$ be a ring and let $J$ be a left ideal of $R$.
Assume that $J$ is nilpotent. Prove that $J$ is contained in a nilpotent 2-sided ideal of $R$.
Comments:
I have found lots of solutions online which go through this proof but can't fully understand any of them. I want to try and use the fact that $J +JR ⊆ JR$.
One proof online stated the following:
"It is easy to see that $J^n = 0$ implies $(J + JR)^n = 0$"
I can't seem to figure out why this would be easy to see.

You need to show that the $2$-sided ideal $J'$ generated by $J$ is nilpotent. For this, it is enough to show that the product of $n$ elements in of the form $jr$, with $j\in J$ and $r\in R$, is zero. This follows from the fact that $J$ is a nilpotent left ideal: if $j_1,\dots,j_n\in J$ and $r_1,\dots,r_n\in R$, then
$$j_1r_1j_2r_2\cdots j_{n-1}r_{n-1}j_nr_n=\underbrace{j_1j'_2\cdots j'_{n-1}j'_n}_{=0}r_n=0$$
with $j_k'=r_{k-1}j_k\in J$ for $k=2,\dots,n$.
