Bertrand's postulate states that for any positive integer n>1 there is a prime with n<p<2n.
The following is not a rigorous proof but rather something I have thought of which makes the theorem intuitively evident, unlike the rigorous proofs which exist (all of which I have seen are very elegant, but tackle the problem in a roundabout fashion).

Suppose we are given an interval I of integers. Call s_p the proportion of integers in I which are multiples of p. Then

s_p\leq \frac{1}{p}
is trivial. In perticular

R(n) = \prod_{p\leq n}\Big(1-\frac{1}{p}\Big)

can be thought of as the proportion of all integers which are not divisible by any of the primes less than n. If we have an interval containing k numbers, we can expect approximately kR(n) of those numbers to be divisible by none of the primes less than n.

Note that we have the ridiculously bad bound

R(n) = \prod_{p\leq n}\Big(1-\frac{1}{p}\Big) \geq \prod_{j=2}^n\Big(1-\frac{1}{j}\Big) = \frac{(n-1)!}{n!}=\frac{1}{n}

so that in perticular any interval containing n numbers should certainly be expected to contain an integer not divisible by any prime less than n, since nR(n)\geq 1. In perticular, when this is applied to the interval ]n,2n], we see that it should very reasonably contain such an integer, which, in this case, would mean a prime.