Let $I$ be an ideal in a Noetherian ring. Show that either $I$ contains an $R$-regular element or else $aI=0$ for some $0\neq a\in R$. Let $I$ be an ideal in a Noetherian ring. Show that either $I$ contains an $R$-regular element or else $aI=0$ for some $0\neq a\in R$.
How would I prove this? Also what does $aI=0$ mean?

Yes this is an important property of Noetherian rings.  It is Theorem 82 in Kaplansky's Commutative Rings, which he prefaces as "a result that is among
the most useful in the theory of commutative rings." 
In the literature you often encounter this property as Property (A)

A ring is said to have Property (A) if every finitely generated dense ideal contains a regular element. (Or equivalently, if every f.g. ideal consisting entirely of zero divisors has a nonzero annihilator.)  

Kaplansky shows that Noetherian rings have Property (A).  
It also comes up in the study of the integral closure of $R[x]$ and the total ring of fractions of a reduced ring.  
