Bertrand's Postulate follows as a direct consequence of the following theorem of J.J. Sylvester:
Theorem (Sylvester, 1892): Let $k$ be a positive integer. Then at least one of any $k$ consecutive integers greater than $k$ is divisible by a prime greater than $k$.
(For comparison: Chebyshev's analytic proof dates to 1850; Erdos' elementary proof dates to 1932.)
See Theorem 6 (p. 6) in http://www.math.sc.edu/~filaseta/papers/schurpaper.pdf, from which I quote:
"The theorem implies immediately that for any positive integer $k$, one of $k+1, k+2, \ldots, 2k$ is a prime (since one of these integers must be divisible by a prime $\geq k+1).$"
Unfortunately, my internet search has not led me to the original paper by Sylvester.