How far can we get?
Here are the rules:
Use four 4's and operators (+,-,*,/,!,^, and square roots, concatenation) to get whole numbers. We can only go in order.
\(0={(4-(\sqrt4}*\sqrt4))^{4!}\)
\(1=\frac{44}{44}\)
See if you can get 2 and then 3 and then 4 and then 5 and so on...
+9464 I am not very good at these.....but here's ones for 2 and 3 and 4:
2 = \(\frac{4*4}{4+4}\)
3 = \(\sqrt{4}+\sqrt4-\frac44\)
4 = \(4! - 4*4 - 4\)
Also...what do you mean by concatenation?
I looked it up but I still don't really know what it is
*edit*
I accidentally put 5 fours for 2...thanks to geno for letting me know!! 
Good job, hectictar!
Concatenation means the action of linking together. For example, 44 is a concatenation, or a link, of a 4 and another 4. Makes sense, no? It's an easier concept to explain than to say the vocab word...
Anyway, here comes the next few.
\(5=\sqrt4+\sqrt4+\frac{4}{4}\)
\(6=\frac{4!}{4+4}*\sqrt4\)
+9464 Ah okay..thanks for the explanation!
Here's a 7, 8, and 9...(...These ones seemed a little too easy 
)
\( 7=4+4-\frac44 \\~\\ 8=4+4*\frac44 \\~\\ 9=4+4+\frac44\)
+2440 \(10=4+4+4-\sqrt4 \)
\(11=\frac{44}{\sqrt4^{\sqrt4}}\)
\(12=\sqrt4(\sqrt4+\sqrt4+\sqrt4)\)
.
+128399 Here's some more (without concatenation )
4! / √4 - 4 / 4 = 24 / 2 - 1 = 12 - 1 = 11
4! / √4 + 4/4 = 24/2 + 1 = 12 + 1 = 13
4! / √4 + 4 / √4 = 24/2 + 4/2 = 12 + 2 = 14
4*4 - 4/4 = 16 - 1 = 15
4 + 4 + 4 + 4 = 16
4 * 4 + 4/4 = 16 + 1 = 17
4 * 4 + 4 / √4 = 16 + 4/2 = 16 + 2 = 18
4! - 4 - 4/4 = 24 - 4 - 1 = 19
4!/√4 + 4 + 4 = 24/2 + 4 + 4 = 12 + 4 + 4 = 12 + 8 = 20
4! - 4 + 4/4 = 24 - 4 + 1 = 20 + 1 = 21
4! - 4 + 4/√4 = 24 - 4 + 4/2 = 24 - 4 + 2 = 20 + 2 = 22
4! - (√4 *√4) / 4 = 24 - (2 *2)/4 = 24 - 4/4 = 24 - 1 = 23
4! - 4/√4 + √4 = 24 - 4/2 + 2 = 24 - 2 + 2 = 22 + 2 = 24
