I also got some interest in the problem of finding, a number between 1 and n^2, which has the highest number of factors (including 1 and itself). One can easily prove that such a number is not divisible by any prime number >= n, for n > 4. Try it.

Of course, this problem can be formalized as optimization problem.
Given that,

k1*log(2) + k2*log(3) + k3*log(5) + .... all prime logs with co-efficients < log(n)
(or, equivalently, 2^k1*3^k2*5^k3*7^k4*..... < n)

maximize (k1 + 1)*(k2 + 1)*(k3 + 1)*(k4 + 1)*....

Finally the number, with most factors is given by 2^k1*3^k2*5^k3*7^k4*.....

Also, I believe n! is not a square number, for any n > 1. Can you prove it? I think it involves the proven conjecture that between any 2n and 3n, there exists a prime number. Can you prove using that statement? Can we do without it?

In fact, I think n! is not any power of any number, not k^2, k^3, k^4 etc... Can you prove this also?

Ok, I give up. I tried doing 101-200 numbers, but, could not get satisfying stuff. Please fill in the gaps.

4!*4 + sqrt(4)/.4
4^4*.4 - .4
103
4!*4 + 4 + 4
(44 - sqrt(4))/.4
4!*4 + 4/.4
107
4!*4 + 4!/sqrt(4)
(44 - .4)/.4
44/.4
(44 + .4)/.4
44/.4 + sqrt(4)
113
44/.4 + 4
(44 + sqrt(4))/.4
4!*sqrt(4)/.4 - 4
117
4!*sqrt(4)/.4 - sqrt(4)
(4!*sqrt(4) - .4)/.4
4!*sqrt(4)/.4
(4!*sqrt(4) + .4)/.4
4!*sqrt(4)/.4 + sqrt(4)
123
4!*sqrt(4)/.4 + 4
(4!*sqrt(4) + sqrt(4))/.4
4^4/sqrt(4) - sqrt(4)
(4^4 - sqrt(4))/sqrt(4)
4^4/sqrt(4)
(4^4 + sqrt(4))/sqrt(4)
4^4/sqrt(4) + sqrt(4)
131
4^4/sqrt(4) + 4
133
44/.4 + 4!
135
4*(4! + 4/.4)
137
(4! - .4)/.4 * sqrt(4)
139
(4^4 + 4!)/sqrt(4)
141
(4! + .4)/.4 * sqrt(4)
143
((4!)^sqrt(4))/4
(4!/.4 - sqrt(4))/.4
4!/.4/.4 - 4
147
4!/.4/.4 - sqrt(4)
(4!/.4 - .4)/.4
4!/.4/.4
(4!/.4 + .4)/.4
4!/.4/.4 + sqrt(4)
153
4!/.4/.4 + 4
(4!/.4 + sqrt(4))/.4
(4!/4)!/4 - 4!
157
158
159
(4!/.4 + 4)/.4
161
162
163
164
165
166
167
4!*(4 + sqrt(4)) + 4!
((4! + sqrt(4))^sqrt(4))/4
(44 + 4!)/.4
171
44*4 - 4
173
((4!/4)! - 4!)/4
(4! + 4)/.4/.4
(4!/4)!/4 - 4
177
(4!/4)!/4 - sqrt(4)
((4!/4)! - 4)/4
(4!/4)!/4
((4!/4)! + 4)/4
(4!/4)!/4 + sqrt(4)
183
(4!/4)!/4 + 4
185
((4!/4)! + 4!)/4
187
4!*(4 + 4) - 4
189
4!*(4 + 4) - sqrt(4)
191
4!*(4 + 4)
193
4!*(4 + 4) + sqrt(4)
195
4!*(4 + 4) + 4
197
198
199
(4!*4 + 4)*sqrt(4)

Do you know, every prime number p > 3, satisfies the following:

p^2 = 24*k + 1, for some number k.
Also, every prime p > 30,
p^4 = 240*k + 1, for some number k.

Try proving them. Also, an extension of Eratoshenes sieve method gives the following approximation of ratio of first n numbers to primes in them:

Number of primes ~= n*(1 - 1/2)(1 - 1/3)(1 - 1/5)(1 - 1/7)(1 - 1/11).....(1 - 1/pk) where pk is the largest prime < sqrt(n).

Ratio is thus, approximately, (1 - 1/2)(1 - 1/3)(1 - 1/5)(1 - 1/7)(1 - 1/11).....(1 - 1/pk)

Also, a composite number (n) definitely has a prime factor <= sqrt(n). Do you know? Try proving it.

Reverted the look back to original one, retro look is gone now.

Look at this:
http://www.cut-the-knot.org/arithmetic/funny/1_4_Func.shtml

Using trignometric functions etc..., makes us able to represent any natural number, using one instance of any other natural number (of course, lots of trig function calls and reciprocal function calls there).

Looks like finding a representation of 75, using only 3 4's is great here. It will solve the issues for representing 73/77 and 99. If we are lucky with the representation of 75, we may be able to do 87 also.

Okey, I could not wait longer, so, browsed the net for other sites with such problems. Here is one, similar problem (not exactly the same, since they use "whole part function", kind of cheating if you ask me), with different approach, and representations:
http://www.cut-the-knot.org/arithmetic/funny/4_4.shtml

For the mathematically inclined, here is something:

We have numbers from 0 to 100 written here, using at max 4 4's and some math operators. The operators that are used are, +, -, *, /, sqrt, . (decimal), ! (factorial) and ^ (to the power of).

However, I could not do some numbers, which are 73/77/87/93 and 99. Interested, please jump in.

(As an aside, is there some theory around these puzzles? I see these in my newspaper every day, and decided to compute these numbers).

4 - 4
4/4
sqrt(4)
sqrt(4) + 4/4
4
4 + 4/4
4 + sqrt(4)
4 + sqrt(4) + 4/4
4 + 4
4 + 4 + 4/4
4 + 4 + sqrt(4)
(4! - sqrt(4))/sqrt(4)
4!/sqrt(4)
(4! + sqrt(4))/sqrt(4)
(4! + 4)/sqrt(4)
4*4 - 4/4
4*4
4*4 + 4/4
4*4 + sqrt(4)
4! - 4 - 4/4
4! - 4
4! - 4 + 4/4
4! - sqrt(4)
4! - 4/4
4!
4! + 4/4
4! + sqrt(4)
4! + sqrt(4) + 4/4
4! + 4
4! + 4 + 4/4
4! + 4 + sqrt(4)
4! + (4! + 4)/4
4! + 4*sqrt(4)
4! + (4 - .4)/.4
4! + 4*sqrt(4) + sqrt(4)
4! + 44/4
4! + 4!/sqrt(4)
4! + (4! + sqrt(4))/sqrt(4)
44 - 4!/4
(4*4 - .4)/.4
44 - 4
(4*4 + .4)/.4
44 - sqrt(4)
44 - 4/4
44
44 + 4/4
44 + sqrt(4)
4!*sqrt(4) - 4/4
4!*sqrt(4)
4!*sqrt(4) + 4/4
4!*sqrt(4) + sqrt(4)
(4! - 4 + .4)/.4
4!*sqrt(4) + 4
(4! - sqrt(4))/.4 - sqrt(4)
(4! - 4)/.4 + 4
(4! - .4)/.4 - 4
4!/.4 - 4
(4! + .4)/.4 - 4
4!/.4 - sqrt(4)
(4! - .4)/.4
4!/.4
(4! + .4)/.4
4!/.4 + sqrt(4)
(4! + .4)/.4 + sqrt(4)
4!/.4 + 4
(4! + .4)/.4 + 4
(4! + 4)/.4 - 4
(4! + sqrt(4))/.4 + sqrt(4)
(4! + 4)/.4 - sqrt(4)
(4! + 4 - .4)/.4
(4! + 4)/.4
(4! + 4 + .4)/.4
(4! + 4)/.4 + sqrt(4)
73 not present
(4! + 4)/.4 + 4
(4! + 4 + sqrt(4))/.4
4!/.4 + 4*4
77 not present
4*(4! - 4) - sqrt(4)
(4! - sqrt(4))/.4 + 4!
(4! + 4 + 4)/.4
(4 - 4/4)^4
4!/.4 - sqrt(4) + 4!
(4! - .4)/.4 + 4!
4!/.4 + 4!
(4! + .4)/.4 + 4!
4!/.4 + sqrt(4) + 4!
87 not present
44*sqrt(4)
(4! + sqrt(4))/.4 + 4!
44*sqrt(4) + sqrt(4)
4!*4 - sqrt(4)/.4
4!*4 - 4
93 not present
(4! + 4)/.4 + 4!
4!*4 - 4/4
4!*4
4!*4 + 4/4
4!*4 + sqrt(4)
99 not present
4!*4 + 4

The scene from 101-200 is not that rosy. Many numbers can't be properly expressed here. That work is un-finished as of now. Will post it soon, as I conclude it.