# Why does the entanglement entropy give the number of singlets required to create a given state?

I've read that, given a bipartite pure state $$|\Phi\rangle$$, its entanglement (equivalently here, von Neumann) entropy $$E(\Phi)$$ gives the asymptotic number of singlets required to create $$n$$ copies of $$|\Phi\rangle$$. More specifically, that for all $$\epsilon>0$$ there is a large enough $$n$$ such that, given $$m$$ copies of a singlet state $$|\Phi^-\rangle$$, Alice and Bob can produce $$n$$ copies of $$|\Phi\rangle$$ with $$m/n\le (1+\epsilon)E(\Phi)$$. I think this roughly means that $$E(\Phi)$$ is the number of singlets required to generate each copy of $$|\Phi\rangle$$.

This is mentioned in this paper by Wooters: http://www.rintonpress.com/journals/qic-1-1/eof2.pdf, around equation 3.

To back up this statement, they cite this paper by Bennett and collaborators. I know that the above statement is shown in this paper, but I was wondering if there was a way to briefly sketch the reasoning, or more generally a way to understand the intuition behind it. More recent takes on this result would also be welcome, if there is any.