1. Prove that 1/64 < (1/2)(3/4)(5/6)... (2009/2010) < 1/44 

2. Find the smallest positive integer N such that N! is a multiple of 10²⁰⁰⁹.

1. Proof that 1/64 < (1/2)(3/4)(5/6)... (2009/2010) < 1/44:

Let P₁ = (1/2) (3/4) (5/6) … (2009/2010) (the product in question)
Let P₂ = (2/3) (4/5) (6/7) … (2010/2011)
Let P = (1/2) (2/3) (3/4) … (2010/2011) = 1/2011

P = P₁ P₂, and P₁ < P₂, so P₁ < sqrt(P) = sqrt(1/2011) < 1/44

Let Q₁ = (2/3) (4/5) (6/7) … (2008/2009)
Let Q₂ = (3/4) (5/6) (7/8) … (2009/2010)
Let Q = (2/3) (3/4) (4/5) … (2009/2010) = 2/2010 = 1/1005

Q = Q₁ Q₂, and Q₂ > Q₁, so Q₂ > sqrt(Q) = sqrt(1/1005) > 1/32
P₁ = (1/2) Q₂ > 1/64, which completes the proof.

The actual value of (1/2)(3/4)(5/6)...(2009/2010) is between 1/57 and 1/56.

2. Find the smallest positive integer N such that N! is a multiple of 10²⁰⁰⁹.

The answer is 8050.

The number of zeros at the end of N! is equal to the exponent of 5 in the prime factorization of N! (because there are plenty of factors of 2, it's the 5's that are in short supply).  The exponent of 5 in N! is [N/5]+[N/25]+…, which equals just under N/4.  (I use square-brackets to denote the "floor" function.)  So if we need 2009 trailing zeros, and 2009 is just under N/4, then N is just over 8036.
Applying the formula, above, we see that 8037! Has 2006 trailing zeros.  The number of trailing zeros doesn't increase until N reaches a multiple of 5, so we can check 8040, and see it has 2007 trailing zeros, 8045 has 2008 trailing zeros, and 8050 has 2010 trailing zeros.

Although there is no number, N, for which N! ends in exactly 2009 zeros, 8050 is the smallest N for which N! ends in at least 2009 zeros.

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