My favourite maths puzzle

This problem, from Peter Winkler's wonderful  Mathematical Puzzles: A Connoisseur's Collection,  is one of my favourites.  At first sight it seemed impossible that there could be a simple solution, but when I looked at the answer I felt that I really should have been able to solve it for myself.

Let t be the positive square root of 2.  Since I'm writing this on 5th May 2012, I will ask "What is the 55th digit after the decimal point of (1+t) to the power 2012?"

Try to solve it before reading on!

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Why does this seem so difficult?  We're raising a irrational number which is a bit more than 2.4 to a very high power and asking for an apparently random digit in the fractional part.  How can we do it?

(1+t)^2012 looks fairly intractable.  We can use the Binomial Theorem to expand it and we will get lots of powers of t.  Even powers are fine, because they are powers of 2, but the odd powers are all irrational and it's hard to see how we can deal with them.

There is a standard trick for getting rid of the irrational numbers.  If we expand (1-t)^2012, then we get exactly the same binomial coefficients but the odd powers have the opposite sign.  So if we were to consider

(1+t)^2012 + (1-t)^2012 (*)

the oddd powers cancel and we are left with the sum of integer multiples of the even powers of t, all of which are integers.  So (*) is an integer.

But how does that help us?

Well, what is (1-t)^2012?  (1-t) is about -0.414, which is less than -1/2. So (1-t) to the power 2012 is less than (-1/2) to the power 2012, which is a very small number indeed.  Since 2^10 is greater than 10^3, (-1/2)^2012 is less than 10^(-670).  So (1-t)^2012 is a very small decimal number beginning with least 670 zeroes.

And so since the second term in the integer (*) is this tiny number beginning with 670 zeroes, so the number we are interested in, being less than an integer by this tiny number, must have a decimal fractional part beginning with at least 670 nines.  So the 55th digit (and another 600-odd after it) must be a 9.

Having seen the solution, one feels that there could have been no other way to tackle this problem.  But I didn't see it for myself.  I've used the problem in a number of Christmas competitions and no-one has solved it yet.  Did you?