② Number the prime numbers from small to large, with 2 being the first prime number, 3 being the second prime number, and so on.

Proof: the nth prime number p (n)

When n= 1, p (1) = 2

Suppose that when n = 1, 2, …, k, all propositions hold.

When n=k+ 1

P(k+ 1)≤P( 1)* P(2)*…* P(k)+ 1 & lt；

& lt2 * P( 1)* P(2)*…* P(k)& lt；

& lt2*[2^(2^ 1)]*[2^(2^2)]*…*[2^(2^k)]=

=2^( 1+2^ 1+2^2+…+2^k)=

=2^[2^(k+ 1)- 1]<；

& lt2 [2 (k+ 1)], the proposition also holds.

P (k+1) ≤ p (1) * p (2) * … * p (k)+1,which is derived from Euclid's infinite prime numbers, can be called Euclid inequality. Bancé inequality is superior to Euclid inequality;

p(k+ 1)& lt； [p (1) * p (2) * ... * p (k)] (1/2), where n≥4.