nilpotent right ideals 
Theorem 3: Every nilpotent right (left) ideal is contained in a nilpotent two-sided ideal.
Proof: Let $I$ be a nilpotent right ideal of $R$.
By induction $(I + RI)^n  ≤ I^n + RI^n$ for all $n≥1$.
so $I + RI$ is a nilpotent two-sided ideal.

I am struggling to understand the proof of this theorem and would appreciate any help in explaining it.
 A: Note that the statement about $(I+RI)^n\subseteq I^n+RI^n$ is valid for every right ideal $I$, not necessarily nilpotent.
The base step ($n=1$) is obvious: $I+RI=I+RI$.
So, assume we know that $(I+RI)^n\subseteq I^n+RI^n$. Then
$$
(I+RI)^{n+1}=(I+RI)^{n}(I+RI)\subseteq(I^n+RI^n)(I+RI)
$$
by the induction hypothesis. Now
$$
(I^n+RI^n)(I+RI)=I^nI+RI^nI+I^nRI+RI^nRI
$$
Since $I^nR\subseteq I^n$, we have $I^nRI\subseteq I^{n+1}$, so also $RI^nRI\subseteq RI^{n+1}$ and we have proved the statement.
If we add the hypothesis that $I$ is nilpotent, then $I^m=0$ for some $m$. Then
$$
(I+RI)^m\subseteq I^m+RI^m=0
$$
and $I+RI$ is nilpotent. It plainly is a two-sided ideal:
A: That $I$ is a right ideal means $IR=I$, so $R(I + RI)=RI\leqslant I+RI$ and $(I+RI)R=I+RI$, thus $I+RI$ is a two-sided ideal. Furthermore, since 
$(I^n + RI^n)(I+RI)=I^{n+1}+RI^{n+1}+I^nRI+RI^nRI=I^{n+1}+RI^{n+1}+I^{n+1}+RI^{n+1}=I^{n+1}+RI^{n+1},$
by induction on $n$ we get $(I + RI)^n =I^n + RI^n$. Finally, if $I^m=0$, then $(I + RI)^m=0$.
