Alternative pattern?

Each number has a larger absolute value but opposite sign. Therefore the sequence becomes larger, smaller, larger, smaller, etc. than all previous numbers. As Shadman said, the original number must be negative for the number to become larger, then smaller, then larger, etc. Also, for the number to be larger (or smaller) than all the previous numbers, the absolute value of the value must increase as each multiplication is performed. This only happens if the absolute value of the original number is greater than 1 , or "number < -1". If the absolute value of the number is less than 1, the value will decrease each time, as in 0.5 x 0.5 = 0.25. Therefore the original number is negative and less than -1 . (E.g. -1.1 x -1.1 = 1.21; 1.21 x -1.1 = -1.331; -1.331 x -1.1 = 1.4641) Here is a completely different solution: $\\ \DeclareMathOperator\min{min}\DeclareMathOperator\max{max}\text{Let the original number be }x.\\ \begin{array}{rl} \therefore x^2&>&x \\ x^3&<&min(x,x^2) \\ x^4&>&max(x,x^2,x^3) \\ x^5&<&min(x,x^2,x^3,x^4) \\ \text{and hopefully }x^6&>&max(x,x^2,x^3,x^4,x^5) \end{array} \\ x^2>x\implies x^2-x>0\implies x(x-1)>0\implies x<0 \text{ or } x>1 \\ x^3<min(x,x^2)\implies x^3<x \text{ (since }x^2>x)\implies x^3-x<0\implies x(x+1)(x-1)<0\implies x<-1 \text{ or }0<x<1 \\ \therefore x<-1 \text{ (combining these intervals with those from the previous line)} \\ x^4>max(x,x^2,x^3)\implies x^4>x^2\implies x^4-x^2>0\implies x^2(x+1)(x-1)>0\implies x<-1 \text{ or }x>1\implies x<-1 \\ x^5<min(x,x^2,x^3,x^4)\implies x^5<x^3\implies x^5-x^3<0\implies x^3(x+1)(x-1)<0\implies x<-1 \text{ or }0<x<1\implies x<-1. \\$ thus proving that only numbers less than -1 will satisfy the conditions. The next in the sequence will be $x^6\boxed{>}max(...)$ .

This problem is solvable only for a negative value of the original number, because multiplication of a positive number by itself would only result in the product becoming larger consecutively. Let the required number be A, where A < -1.

Hence, Multiplication of A with itself = A multiplied by A = A^(2)

Here this number is positive because multiplication of two negative quantities always give a positive product, i.e. (-a) multiplied by (-b) = ab

Again, multiplication of A^2 with A ( the original number) = A^(2) multiplied by A = A^(3).

In this case, A^(2) was positive, but A was negative. Multiplication of one negative and one positive number will always give a negative number, i.e. (a) multiplied by (-b) = -ab

So, for any odd number of times A is multiplied by itself, the product will be negative, hence smaller than all the previous numbers. Again, for any even number of times A is multiplied by itself, the product will be positive, hence larger than all the previous numbers.

Therefore, when A was multiplied by itself 6 times, the product was larger than the previous 5.