The idea is quite well-known. The set of the first $2011^{2011}$ consecutive positive integers contains $p> 2011$ primes (this is probably where that piece of information about the $2011$-th prime is used; however, by Bertrand's postulate, $p_n \leq 2^n$, so $p_{2011} < 2^{2011} < 2011^{2011}$, so we could do without that). On the other hand, for any $N$ there exist $N$ consecutive composite positive integers, for example $(N+1)! + 2, (N+1)! + 3, (N+1)! + N + 1$. By something sometimes called discrete continuity, since the number of primes in a sequence of $2011^{2011}$ consecutive positive integers changes by at most $\pm 1$ as we shift the sequence by one number to the right, it follows that when passing from $p$ to $0$ it must be at some moment that the number of primes is precisely $2011$.