Ideal Generated by Nilpotent Elements is a Nilpotent Ideal 
Prove that if $R$ is a commutative ring and $N=(a_1,...,a_m)$ where each $a_i$ is nilpotent., then $N$ is a nilpotent ideal. 

So we need to find a $k \in \Bbb{N}$ such that $N^k = (0)$. Working through small values $m$ and $k$, I can see that $k = 1 + \max \{a_1,...,a_m\}$ seems to work. E.g., if $m=2$ and $k=3$, then 
$$(a_1,a_2)^3 =[(a_1)+(a_2)]^3 \subseteq (a_1)^3 + (a_1)^2(a_2) + (a_1)(a_2)^2 + (a_2)^3 = (0),$$
which is true because the nilpotency of each element is no larger than $2$. However, I can't get the general case to work: I've tried induction on $m$ and induction on $k$, but neither panned out. I could use a hint.
 A: The statement is obvious when $m=1$. Now suppose you have $N=(a_1,\dots,a_m,a_{m+1})$, with $a_i$ nilpotent.
By the induction hypothesis, $I=(a_1,\dots,a_m)$ is nilpotent, so $I^r=(0)$ for some $r$. Set $a=a_{m+1}$ for simplicity, with $a^s=0$. You know that $N=I+Ra$. Consider
$$
(I+Ra)^{r+s}
$$
Start with a general identity for $(I+J)^n$, where $I$ and $J$ are ideals of $R$.

 You should be able to prove that $$(I+J)^n=\sum_{k=0}^n I^kJ^{n-k}$$

A: Hint: try $k = m\times \max\{a_1, \dots, a_m\}$
A: Excuse me for being verbose.
Suppose $I$ is generated by two nilpotent elements $a,b$ so that $a^k=0$ and $b^n=0$, with $k,n>0$.
We want to ensure that no matter which way we multiply some $N$ elements, we always get zero for some $N \in \mathbb N$.
Suppose wlog that $n>k$.
Well, if $N=2 \cdot n $ for example, then the pigeonhole principle tells that there are $2n$ "slots", and only $2$ "pigeons." In other words, there must be $n$ of one of the choices, so either $a^n$ is in the product, or $b^n$ is in the product. In either case, the product is zero.
The point here is that $I^{k}$ consists of elements that are $k$ products of elements in the ideal. It's enough to check whether or not the generators are $0$ after a $k$-fold product, and that will just be the case for $k$ sufficiently large, so all of th elements are zero.
