In 1949, Leo Moser gave a proof of  Bertrand’s postulate in the paper “A theorem of the distribution of primes”(American Mathematical Monthly,624-624,1949,vol56). Maybe this proof  is the simplest one among all proofs of  Bertrand’s postulate.

1. If $$n<p\leqslant 2n$$, then $$p$$ occurs exactly once in $${2n\choose n}$$;
2. If $$n\geqslant 3, \frac{2n}3<p<n$$, then $$p$$ doesn’t occurs in $${2n\choose n}$$;
3. If $$p^2>2n$$, then $$p$$ occurs at most once in $${2n\choose n}$$;
4. If $$2^\alpha\leqslant 2n<2^{\alpha+1}$$, then $$p$$ occurs at most $$\alpha$$ times in $${2n\choose n}$$.

If  $$2^\alpha<2n\leqslant 2^{\alpha+1}$$, suppose there is no prime $$p$$ with

$$n<p<2n,$$

then

$${2n\choose n}\leqslant{2a_1\choose a_1}{2a_2\choose a_2}\dotsm{2\choose 1}\bigg({2a_k\choose a_k}{2a_{k+1}\choose a_{k+1}}\dotsm{2\choose 1}\bigg)^\alpha,$$

where $$a_1=\big[\frac{n+1}3\big], k=\big[\frac{\alpha+1}2\big]$$, and $$a_i=\big[\frac{a_{i-1}+1}3\big]$$, for $$i=2,\,3,\,\dotsc\,.$$

By $$1,\,2,\,3$$ and $$(1)$$, every prime which appears on the left-hand side of $$(2)$$ appears also on the right; and those primes which appear with multiplicity greater than $$1$$ on the left appear on the right with multiplicity at least $$2\alpha+1$$, which by $$4$$ is at least equal to the multiplicity with which they appear on the left.

On the other hand, It is obvious that

$$2n>2a_1+2a_2+\dotsb+ 2+\alpha(2a_k+2a_{k+1}+\dotsb+2)$$

for $$n>2^{11}.$$ Hence the inequality in $$(2)$$ should be reversed. So for $$n>2^{11}$$, we have a contradiction which proves Bertrand’s postulate for these values of $$n$$.

Sylvester’s theorem  the binomial coefficient

$${n\choose k}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}$$

has a prime divisor $$p$$ greater than $$k$$. In other words, if $$n\geq 2k$$, then the product of  $$k$$ consecutive integers greater than $$k$$ is divisible by a prime greater than $$k$$. Note that for $$n=2k$$ we obtain precisely Bertrand’s postulate.

Theorem

1. if $$n>6$$ is an integer, then there always exists at least two prime numbers $$p$$, one  is the form $$4k+1$$ and the other $$4k+3$$, such that $$n<p<2n$$.
2. if $$k$$ is a positive  integer, then there always exists a positive  integer $$N$$ such that  for every  integer $$n>N$$,  there exist at least $$k$$ prime numbers $$p$$ such that $$n<p<2n$$.
• 对任意正整数 $$n>6$$ ,  至少存在一个 $$4k+1$$ 型和一个 $$4k+3$$  型素数 $$p$$  使得  $$n<p<2n$$.
• 对任意正整数 $$k$$,  存在正整数 $$N$$,  使得只要正整数 $$n>N$$,  就存在 $$k$$ 个素数 $$p$$ 使得 $$n<p<2n$$.