[avian]

No. If pi would contain itself it would be a rational number. We have a proof that pi is irrational, hence it cannot contain itself.

[gowld]

Specifically, if pi contains itself, that means "pi - pi/b^k" is a terminating fraction N in base b", where b is your choice of base (like base 10), and k is some positive integer. So

pi = N/(1-b^k) which is rational.

(And if you set k=0, you get that pi "contains itself" in the sense that pi from the 1st digit onward is of course pi)