The motivation for this question comes from the novel Contact by Carl Sagan.  Actually, I haven't read the book myself.  However, I heard that one of the characters (possibly one of those aliens at the end) says that if humans compute enough digits of $\pi$, they will discover that after some point there is nothing but zeroes for a really long time.  After this long string of zeroes, the digits are no longer random, and there is some secret message embedded in them.  This was supposed to be a justification of why humans have 10 fingers and increasing computing power.  

Anyway, apologies for the sidebar, but this all seemed rather dubious to me.  So, I was wondering if it is known that $\pi$ does not contain 1000 consecutive zeroes in its base 10 expansion?  Or perhaps it does?  Of course, this question makes sense for any base and digit.  Let's restrict ourselves to base 10.  If $\pi$ does contain 1000 consecutive $k$'s, then we can instead ask if the number of consecutive $k$'s is bounded by a constant $b_k$.

According to the wikipedia page, it is not even known which digits occur infinitely often in $\pi$, although it is conjectured that $\pi$ is a normal number.  So, it is theoretically possible that only two digits occur infinitely often, in which case $b_k$ certainly exist for at least 8 values of $k$.    

Update. As Wadim Zudilin points out, the answer is conjectured to be yes.  It in fact follows from the definition of a normal number (it helps to know the correct definition of things).  I am guessing that a string of 1000 zeroes has not yet been observed in the over 1 trillion digits of $\pi$ thus computed, so I am adding the open problem tag to the question.  Also, Douglas Zare has pointed out that in the novel, the actual culprit in question is a string of 0s and 1s arranged in a circle in the base 11 expansion of $\pi$.  See here for more details.  

Summing up what others have written, it is widely believed (but not proved) that every finite string of digits occurs in the decimal expansion of pi, and furthermore occurs, in the long run, "as often as it should," and furthermore that the analogous statement is true for expansion in base b for b = 2, 3, .... On the other hand, for all we are able to prove, pi in decimal could be all sixes and sevens (say) from some point on. 

About the only thing we can prove is that it can't have a huge string of zeros too early. This comes from irrationality measures for pi which are inequalities of the form 
$|\pi-(p/q)|>q^{-9}$ (see, e.g., Masayoshi Hata, Rational approximations to $\pi$ and some other numbers, Acta Arith. 63 (1993), no. 4, 335-349, MR1218461 (94e:11082)), which tell us that such a string of zeros would result in an impossibly good rational approximation to pi. 
