You don' need Bertrand's Postulate to prove it, it is enough with Euler's theorem on the divergence of $\sum \frac{1}{p}$. The argument is as follows:
For some integer $n$, consider the $n-2$ differences $p_{i+1}-p_i$ for $i=2,3,\dots, n-1$, if they take at most $ T_n = \left\lfloor\frac{n-2}{N}\right\rfloor $ different values then there is at least one taken $N$ times and we are done.
Suppose otherwise that for every $n$ there are more than $T_n$ different values in the set $\{ p_{i+1}-p_i\,\,; i=2,3,\dots,n\}$ then $ p_n - p_2 = \sum_{i=2}^{n-1} p_{i+1}-p_i \ge 2+4+6+\dots+2T_n+2(T_n+1) = (T_n+1)(T_n+2) $ but $ (T_n+1)(T_n+2) \ge \frac{(n-2)^2}{N^2}$ So $ \frac{1}{p_n} \le \frac{N^2}{(n-2)^2 + 3N^2} $ if we fix $N$ and sum for $n=N,N+1,\dots,M $ we get $\sum_{i\le M} \frac{1}{p_i} \le -\sum_{i\le N} \frac{1}{p_i} + N^2 \sum_{N\le n \le N} \frac{1}{(n-2)^2 + 3N^2} $ But the right hand side is bounded and by Euler's theorem the right hand isn't so for $n$ large enough we get a contradiction.
Note that this proves something stronger: for every $N$ there is an integer $k$ such that there are more than $N$ consecutive primes with difference $k$.
I read this argument in some old issue of the American Math Monthly (in 1985 or earlier), I'm sorry I can't give a reference.