The number is 27. (33)
The sequence shows the lowest 7 nonzero values for the exponential series nn,
i.e. 1 to the first power, 2 to the 2nd power, up to 7 to the 7th power.
1*1=1
2*2=4
3*3*3=27
4*4*4*4=256
5*5*5*5*5=3125
6*6*6*6*6*6=46656
7*7*7*7*7*7*7=823543
27. The pattern is: 823543 = 6 raised to the 6th 46656 = 5 raised to the 5th ... 27 = 3 raised to the 3rd.
46656 as it is 6 to the power of 6. Each in turn is the result of a number to the power of itself.
46656 = (2^6)*(3^6)
3125, 46656 This is the sequence I used: * 1 = 11 * 4 = 22 * 27 = 33 * 256 = 44 * 3125 = 55 * 46,656 = 66 * and so on...
46656
27. The pattern is: 823543 = 6 raised to the 6th 46656 = 5 raised to the 5th ... 27 = 3 raised to the 3rd.
27
46656 as it is 6 to the power of 6. Each in turn is the result of a number to the power of itself.
46656^7 = 481229803398374426442198455156736
46656 = (2^6)*(3^6)
Yes - there is an infinite number of such numbers. Just take the counting numbers (1,2,3,4, etc...) and put them to the power of 6 (i.e. multiply it by itself 6 times). The sequence of such numbers starts: 1, 64, 729, 4096, 15625, 46656, etc...
It is 36 because 363 = 46656
3125, 46656 This is the sequence I used: * 1 = 11 * 4 = 22 * 27 = 33 * 256 = 44 * 3125 = 55 * 46,656 = 66 * and so on...
46656
It would be 46656.
As a product of its prime factors in exponents: 2^6 times 3^6 = 46656
46656