There are various proofs of Bertrand's postulate. There is quite an easy one available if one treats it together with the proof of the usual (double) Chebyshev bound as a unit. One optimizes the proof of the Chebyshev bound for subsequently proving Bertrand's Postulate by the method of Ramanujan (getting rid of the appeal to Stirling's formula at the same time).

The history of Bertrand's Postulate is set forth in The Development of Prime Number Theory by Wladyslaw Narkiewicz.

A comment to the comment by Michael Lugo. The Prime Number Theorem is considerably harder to prove than Bertrand's Postulate, and getting the PNT in the form of good explicit inequalities is hard work on top of that (such inequalities exist, and are useful for some purposes).

There are various proofs of Bertrand's postulate. There is quite an easy one available as a follow-up to an easy version if one treats it together with the proof of the usual (double) Chebyshev bound , which I give below in tex. You need to know that \vartheta(x) = \sum_{p \leq x}\log(p)\psi(x) = \sum_{p^k \leq x}\log(p)and that T(x) = \sum_{n \leq x}\log(n)Here is the Chebyshev bound...

\begin{Chebyshev} The inequalities \log(2)x - \log(4x) \leq \psi(x) \leq 2\log(2)x + 2\log(x) hold for $x \geq 1$.\end{Chebyshev}

\begin{proof} The terms in the last sum in the computation \begin{align} T(x) - 2T\left(\frac{x}{2}\right) &=
\sum_{n \leq x}\log(n) - 2\sum_{m \leq x/2}\log(m) \ &= \sum_{n \leq x}\log(n) - 2\sum_{2m \leq x}\log(2m) + 2\sum_{2m \leq x}\log(2) \ &= \sum_{n \leq x}(-1)^{n-1}\log(n) + 2\left[\frac{x}{2}\right]\log(2) \end{align} alternate in sign and increase in magnitude. So \left|T(x) - 2\,T\left(\frac{x}{2}\right) - 2\left[\frac{x}{2}\right]\log(2)\right| \leq \log([x]) for $x \geq 1$. Thus \log(2)x - \log(4x) \leq T(x) - 2\,T\left(\frac{x}{2}\right) \leq
\log(2)x + \log(x). Substituting the expression T(x) = \sum_{n \leq x}\psi\left(\frac{x}{n}\right) into $T(x) - 2\,T(x/2)$ yields \psi(x) - \psi\left(\frac{x}{2}\right) + \psi\left(\frac{x}{3}\right) - \cdots = T(x) - 2\,T\left(\frac{x}{2}\right). Then \psi(x) \geq \log(2)x - \log(4x) since $\psi$ is an increasing and nonnegative function. And \psi(x) - \psi\left(\frac{x}{2}\right) \leq \log(2)x + \log(x) for the same reason. Adding up all the inequalities \psi\left(\frac{x}{2^j}\right) - \psi\left(\frac{x}{2^{j+1}}\right) \leq \log(2)2^{-j}x + \log\left(\frac{x}{2^j}\right) for $j = 0,1,2,\ldots,[\log(x)/\log(2)]$ gives \psi(x) \leq 2\log(2)x + \left[\frac{\log(x)}{\log(2)}\right]\log(x) \leq 2\log(2)x + \frac{\log^2(x)}{\log(2)} since $\psi(x/2^{j+1}) = 0$ when $x/2^{j+1} < 1$. Now assume that the inequality \psi(x) \leq 2\log(2)x + \frac{\log^2(x)}{\log(2)} - (k-1)\log\left(\frac{x}{2}\right) holds for as a positive integer $k$. It certainly holds for $k = 1$unit. Apply it on One optimizes the right hand side of \psi(x) \leq \psi\left(\frac{x}{2}\right) + \log(2)x + \log(x) to obtain \begin{align} \psi(x) &\leq 2\log(2)\frac{x}{2} + \frac{\log^2(x/2)}{\log(2)} - (k-1)\log\left(\frac{x}{2}\right) + \log(2)x + \log(x) \ &= 2\log(2)x + \frac{\log^2(x)}{\log(2)} - k\log\left(\frac{x}{2}\right). \end{align} Then \begin{align} \psi(x) &\leq 2\log(2)x + \frac{\log^2(x)}{\log(2)} - \left[\frac{\log(x)}{\log(2)}\right] \log\left(\frac{x}{2}\right) \ &\leq 2\log(2)x + 2\log(x) \end{align} holds by repeated halving.\end{proof}

The advantages proof of this version are two: It does not require Stirling's formula, and it is just right the Chebyshev bound for subsequently proving Bertrand's Postulate by the method of Ramanujan . Here goes...

\begin{Bertrand} For every $x \geq 2$ there exists at least one prime $p$ with $x/2 < p \leq x$.\end{Bertrand}

\begin{proof} The basic idea is to use the difference $\vartheta(x) - \vartheta(x/2)$ to detect primes in the interval $(x/2,x]$. A suitable positive lower bound for $\vartheta(x) - \vartheta(x/2)$ is going to imply that there is at least one prime in $(x/2,x]$ for all $x$ sufficiently large. At the end, the intervals $(x/2,x]$ for small values of $x$ have to be treated separately.

The definitions of $\psi$ and $\vartheta$ yield     \psi(x) &= \sum_{p^k \leq x}\log(p)     = \sum_{k=1}^{\infty}\sum_{p^k \leq x}\log(p) \\    &= \sum_{k=1}^{\infty}\sum_{p \leq x^{1/k}}\log(p) =     \vartheta(x) + \vartheta(x^{1/2}) + \vartheta(x^{1/3})     + \cdotsand so \psi(x) - 2\psi(x^{1/2}) = \vartheta(x) - \vartheta(x^{1/2}) + \vartheta(x^{1/3}) - \cdots.The inequality $\psi(x) - 2\psi(x^{1/2}) \leq \vartheta(x) \leq \psi(x)$ follows since $\vartheta$ is an increasing and nonnegative function.Fetch the inequality \psi(x) - \psi\left(\frac{x}{2}\right) + \psi\left(\frac{x}{3}\right) \geq T(x) - from the proof of the Chebyshev bound. Retention (getting rid of the term $\psi(x/3)$ is an idea due to Ramanujan. Now      \vartheta(x) - \vartheta\left(\frac{x}{2}\right) &\geq     \psi(x) - 2\psi(x^{1/2}) - \psi\left(\frac{x}{2}\right) \\    &\geq T(x) - 2T\left(\frac{x}{2}\right) -     \psi\left(\frac{x}{3}\right) - 2\psi(x^{1/2}) \\    &\geq \log(2)x - \log(4x) - 2\log(2)\frac{x}{3} \\    &- 2\log\left(\frac{x}{3}\right) - 4\log(2)x^{1/2} -     4\log(x^{1/2})by the Chebyshev bound and its proof. So $\vartheta(x) - \vartheta(x/2) \geq f(x)$ with f(x) = \log(2)x/3 - 4\log(2)x^{1/2} - 5\log(x) + The derivative f'(x) = \log(2)/3 - 2\log(2)x^{-1/2} - 5x^{-1}is visibly an increasing function. Then $f'(x) > 0$ on the interval $[347,\infty)$ since $f'(347) = 0.14$. Thus $f(x)$ is increasing on this interval and hence positive there because $f(347) = 0.09$. So there is a prime in every interval $(x/2,x]$ with $x \geq 347$. Now a trick due appeal to E. G. H. Landau     completes the proof: Each term in Stirling's formula at the chain $2,3,5,7,11,17,29,53,97,179,347$ of primes is smaller than twice its predecessor. So there is also a prime in each interval $(x/2,x]$ for $2 \leq x \leq 347$same time).

\end{proof}

There are various proofs of Bertrand's postulate. There is quite an easy one available as a follow-up to an easy version of the Chebyshev bound, which I give below in tex. You need to know that [ \vartheta(x) = \sum_{p \leq x}\log(p) ] and that [ \psi(x) = \sum_{p^k \leq x}\log(p) ] and that [ T(x) = \sum_{n \leq x}\log(n) ] Here is the Chebyshev bound...

\begin{Chebyshev} The inequalities [ \log(2)x - \log(4x) \leq \psi(x) \leq 2\log(2)x + 2\log(x) ] hold for $x \geq 1$. \end{Chebyshev}

\begin{proof} The terms in the last sum in the computation \begin{align} T(x) - 2T\left(\frac{x}{2}\right) &=
\sum_{n \leq x}\log(n) - 2\sum_{m \leq x/2}\log(m) \ &= \sum_{n \leq x}\log(n) - 2\sum_{2m \leq x}\log(2m) + 2\sum_{2m \leq x}\log(2) \ &= \sum_{n \leq x}(-1)^{n-1}\log(n) + 2\left[\frac{x}{2}\right]\log(2) \end{align} alternate in sign and increase in magnitude. So [ \left|T(x) - 2\,T\left(\frac{x}{2}\right) - 2\left[\frac{x}{2}\right]\log(2)\right| \leq \log([x]) ] for $x \geq 1$. Thus [ \log(2)x - \log(4x) \leq T(x) - 2\,T\left(\frac{x}{2}\right) \leq
\log(2)x + \log(x). ] Substituting the expression [ T(x) = \sum_{n \leq x}\psi\left(\frac{x}{n}\right) ] into $T(x) - 2\,T(x/2)$ yields [ \psi(x) - \psi\left(\frac{x}{2}\right) + \psi\left(\frac{x}{3}\right) - \cdots = T(x) - 2\,T\left(\frac{x}{2}\right). ] Then [ \psi(x) \geq \log(2)x - \log(4x) ] since $\psi$ is an increasing and nonnegative function. And [ \psi(x) - \psi\left(\frac{x}{2}\right) \leq \log(2)x + \log(x) ] for the same reason. Adding up all the inequalities [ \psi\left(\frac{x}{2^j}\right) - \psi\left(\frac{x}{2^{j+1}}\right) \leq \log(2)2^{-j}x + \log\left(\frac{x}{2^j}\right) ] for $j = 0,1,2,\ldots,[\log(x)/\log(2)]$ gives [ \psi(x) \leq 2\log(2)x + \left[\frac{\log(x)}{\log(2)}\right]\log(x) \leq 2\log(2)x + \frac{\log^2(x)}{\log(2)} ] since $\psi(x/2^{j+1}) = 0$ when $x/2^{j+1} < 1$. Now assume that the inequality [ \psi(x) \leq 2\log(2)x + \frac{\log^2(x)}{\log(2)} - (k-1)\log\left(\frac{x}{2}\right) ] holds for a positive integer $k$. It certainly holds for $k = 1$. Apply it on the right hand side of [ \psi(x) \leq \psi\left(\frac{x}{2}\right) + \log(2)x + \log(x) ] to obtain \begin{align} \psi(x) &\leq 2\log(2)\frac{x}{2} + \frac{\log^2(x/2)}{\log(2)} - (k-1)\log\left(\frac{x}{2}\right) + \log(2)x + \log(x) \ &= 2\log(2)x + \frac{\log^2(x)}{\log(2)} - k\log\left(\frac{x}{2}\right). \end{align} Then \begin{align} \psi(x) &\leq 2\log(2)x + \frac{\log^2(x)}{\log(2)} - \left[\frac{\log(x)}{\log(2)}\right] \log\left(\frac{x}{2}\right) \ &\leq 2\log(2)x + 2\log(x) \end{align} holds by repeated halving. \end{proof}

The advantages of this version are two: It does not require Stirling's formula, and it is just right for proving Bertrand's Postulate by the method of Ramanujan. Here goes...

\begin{Bertrand} For every $x \geq 2$ there exists at least one prime $p$ with $x/2 < p \leq x$. \end{Bertrand}

\begin{proof} The basic idea is to use the difference $\vartheta(x) - \vartheta(x/2)$ to detect primes in the interval $(x/2,x]$. A suitable positive lower bound for $\vartheta(x) - \vartheta(x/2)$ is going to imply that there is at least one prime in $(x/2,x]$ for all $x$ sufficiently large. At the end, the intervals $(x/2,x]$ for small values of $x$ have to be treated separately.

The definitions of $\psi$ and $\vartheta$ yield
\begin{align}
\psi(x) &= \sum_{p^k \leq x}\log(p)
= \sum_{k=1}^{\infty}\sum_{p^k \leq x}\log(p) \\
&= \sum_{k=1}^{\infty}\sum_{p \leq x^{1/k}}\log(p) =
\vartheta(x) + \vartheta(x^{1/2}) + \vartheta(x^{1/3})
+ \cdots
\end{align}
and so
$\psi(x) - 2\psi(x^{1/2}) = \vartheta(x) - \vartheta(x^{1/2}) + \vartheta(x^{1/3}) - \cdots.$
The inequality
$\psi(x) - 2\psi(x^{1/2}) \leq \vartheta(x) \leq \psi(x)$
follows since $\vartheta$ is an increasing and
nonnegative function.

Fetch the inequality
$\psi(x) - \psi\left(\frac{x}{2}\right) + \psi\left(\frac{x}{3}\right) \geq T(x) - 2T\left(\frac{x}{2}\right)$
from the proof of the Chebyshev bound. Retention of the term
$\psi(x/3)$
is an idea due to Ramanujan. Now
\begin{align}
\vartheta(x) - \vartheta\left(\frac{x}{2}\right) &\geq
\psi(x) - 2\psi(x^{1/2}) - \psi\left(\frac{x}{2}\right) \\
&\geq T(x) - 2T\left(\frac{x}{2}\right) -
\psi\left(\frac{x}{3}\right) - 2\psi(x^{1/2}) \\
&\geq \log(2)x - \log(4x) - 2\log(2)\frac{x}{3} \\
&- 2\log\left(\frac{x}{3}\right) - 4\log(2)x^{1/2} -
4\log(x^{1/2})
\end{align}
by the Chebyshev bound and its proof. So $\vartheta(x) - \vartheta(x/2) \geq f(x)$ with
$f(x) = \log(2)x/3 - 4\log(2)x^{1/2} - 5\log(x) + \log(9/4).$
The derivative
$f'(x) = \log(2)/3 - 2\log(2)x^{-1/2} - 5x^{-1}$
is visibly an increasing function. Then $f'(x) > 0$ on the interval
$[347,\infty)$ since $f'(347) = 0.14$. Thus $f(x)$ is
increasing on this interval and hence positive there because
$f(347) = 0.09$. So there is a prime in every interval
$(x/2,x]$ with $x \geq 347$. Now a trick due to E. G. H. Landau
completes the proof:
Each term in the chain $2,3,5,7,11,17,29,53,97,179,347$ of
primes is
smaller than twice its predecessor. So there is also a prime
in each
interval $(x/2,x]$ for $2 \leq x \leq 347$.


\end{proof}