We begin with the following lemma.
Proof. Let
divide 
with

as large as possible. Every

th term of the
sequence 
is
divisible by

, so contributes a factor to

. There are

such factors. Every

th term contributes an extra factor above that, providing

new factors. In general, the

th terms contribute

extra factors to

. So the highest power of

dividing

is

.

We now prove the theorem.
Proof. [Bertrand's conjecture] Let

be a minimal
counterexample to the claim. Thus there is no prime

such that

.
In the sequence of primes
each succeeding term is smaller than the double of its predecessor. This shows that

.
The binomial expansion of
has
terms and has largest term
. Hence
For a prime
define
to be the highest power of
dividing
. To compute
, we apply the lemma to
and
. We get that
The terms of the
sum are all 0 or 1 and
vanish when

, so

, that is,

.
Now
. By the inequality just proved, primes larger than
do not contribute to this product, and by assumption there are no primes between
and
. So
For
,
and so for
we can apply the previous formula for
and find that it is zero. So for all
, the contribution of the primes larger than
is zero.
If
, all the terms for higher powers of
vanish and
. Since
is at most 1, an upper bound for the contribution for the primes between
and
is the product of all primes smaller than
. This product is
, where
is the Chebyshev function
There are at most

primes smaller than

and by the inequality

their product is less than

. Combining this information, we get the inequality
Taking
logarithms and applying the
upper bound of
for

, we obtain the inequality

, which is false for sufficiently large

, say

. This shows that

.
Since the conditions
and
are incompatible, there are no counterexamples to the claim. 