proof of infinitude of primes
Then, because is the smallest prime number, we must have
Assume that there were only a finite number of prime numbers . By the above-noted fact, given a positive integer , every integer n such that could be expressed as
This, however, leads to a contradiction because it would imply that there exist more integers than possible factorizations despite the fact that every integer is supposed to have a prime factorization. To see this, let us over-count the number of factorizations. A factorization being specified by an -tuplet of integers such that , the number of factorizations is equal to the number of such tuplets. Now, for all we must have , so there are not more than such tuplets available. However, for all , one can choose such that . For such a choice of we could not make ends meet — there are not enough possible factorizations available to handle all integers, so we conclude that there must be more than primes for any integer , i.e. that the number of primes is infinite.
To make this exposition self-contained, we conclude with a proof that, for every , there exists a such that . We begin by showing that, for every integer , we have . This is an easy induction. When , we have . If for some , then . Hence, by induction, for all .
From this starting point, we obtain the desired inequality by algebraic manipulation. Suppose that . Multiplying both sides by , we get . Setting , we have , or . Raising both sides to the -th power, . Setting , this becomes .