Show that there exists $n$ such that $i^n = 0$ for all $i$ in the ideal $I$ I'm new to this medium, but I'm quite stuck with an exercise so hopefully someone here can help me. 
This is the exercise: 
Let $I$ be an ideal in a Noetherian ring $A$, and assume that for every $i\in I$ there exists an $n_i$ such that $i^{n_i} = 0$. Show that there is an $n$ such that $i^n = 0$ for every $i\in I$. 
I thought about this:
$A$ is Noetherian so $I$ is finitely generated. That means, there exist $i_1, \ldots, i_m$ such that all of the elements in $I$ are linear combinations of $i_1, \ldots, i_m$. 
Now is it possible to take $n = n_1n_2\cdots n_m$? 
I was thinking, maybe I have to use something like this: $(a + b)^p = a^p + b^p$. This holds in a field of characteristic $p$. If this is true, then indeed $(a_1i_1 + \ldots + a_mi_m)^n = 0 + \ldots + 0 = 0.$

Pritam's binomial approach is the easiest way to solve this problem in the commutative setting. In case you're interested though, it's also true in the noncommutative setting.
Levitzky's theorem says that any nil right ideal in a right Noetherian ring is nilpotent. This implies your conclusion (in fact, even more.) However this is not as elementary as the binomial proof in the commutative case :)
