Calculator Killer 1

$2!^{8!}$ is the largest of this set.

How to quickly deduce this fact (with absolutely no use of a calculator):

Use a little wishful thinking... If all the bases were the same (and all bases and exponents >1), then whichever option had the biggest exponent would be biggest. Similarly, if all of the exponents were the same, which ever base was largest would be the biggest option. Our goal is to compare pairs from among these 6 optios in one of those two ways -- the two items being compared should either be different bases to the same power , or different powers of the same base . Here is an example of the former technique:

Let's compare $\color{#20A900}{2!^{8!}}$ , and $\color{#D61F06}{5!^{5!}}$ , by converting the former to also have the form $X^{5!}$ :

$2!^{8!} = 2^{(8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}$

Therefore, we can use the rule of exponents , $x^{(a \times b)} = (x^a)^b$ to change the form of the 2! option:

$2!^{8!} = (2 ^{(8 \times 7 \times 6)})^{(5 \times 4 \times 3 \times 2 \times 1)}= (2 ^{(8 \times 7 \times 6)})^{(5!)}$

The comparison is now much clearer: $\color{#20A900}{(2 ^{(8 \times 7 \times 6)})^{5!}}$ vs. $\color{#D61F06}{5!^{5!}}$

Since $2 ^{(8 \times 7 \times 6)} >> 5 \times 4 \times 3 \times 2 \times 1$ it follows that, $2!^{8!} > 5!^{5!}$

In summary...

• $2 ^{(8 \times 7 \times 6)} > 5!$ therefore $2!^{8!} > 5!^{5!}$
• $2 ^{(8 \times 7)} > 4!$ therefore $2!^{8!} > 4!^{6!}$
• $2 ^{8} > 3!$ therefore $2!^{8!} > 3!^{7!}$
• $1!^{9!} = 1$ therefore $2!^{8!} > 1!^{9!}$

2!^8!=2^40320 = 2^(8x5040) =256^5040 Largest!!
3!^7!=6^5040 = 6^(3x1680) = 216 ^1682
4!^6!=24^720
5!^5! = 120^120

Convert each number into the powers of 5!. Like 1!^ 9.8.7.6.5!=1^5!, similarly 2!^8.7.6.5!=2^336.5!= solve it. Similarly othrrs. We obsrv that 2^8! Isargest

I was guessing, but exponential growth is my choice for getting big

1st, we remove $1!^{9!}$ because it equals 1. Then, another 4 answer equals $(2!^{6 \times 7 \times 8})^{5!}$ , $(3!^{6 \times 7})^{5!}$ , $(4!^5)^{5!}$ , $5!^{5!}$ So $2!^{8!}$ is biggest.

I do not know if this is correct reasoning: Someone let me know please. I compared $2!^{8!}$ to $3!^{7!}$ . I wrote them down like $2^{8!}$ and $6^{7!}$ . Then I compared the exponents: 8! is $8$ times larger than 7!. So I reduced to $2^{8}>6^{1}$ . Again, please let me know if this is incorrect.

Take all the bases to a common number divide the factorial according to the previous one starting from 5! To the power 5! And u will get it easily!!