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.

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 $\geq k+1).$"

A copy of the Sylvester paper can be found here.

Edit: An article in the AMM (Aug/Sept, 2013) presents a revised version of Ramanujan's proof of Bertrand's Postulate; in particular, in which the use of Stirling's formula is eliminated. The citation is:

Ramanujan’s Proof of Bertrand’s Postulate. Jaban Meher, M. Ram Murty. The American Mathematical Monthly, Vol. 120, No. 7 (August–September 2013), pp. 650-653. http://www.jstor.org/stable/10.4169/amer.math.monthly.120.07.650.