Prove that the number 14641 is the fourth power of an integer in any base greater than 6? Prove that the number $14641$ is the fourth power of an integer in any base greater than $6$?
I understand how to work it out, because I think you do 
$$14641\ (\text{base }a > 6) = a^4+4a^3+6a^2+4a+1= (a+1)^4$$
But I can't understand why they had to specify that the base is greater than 6? Is that because if it's 4, then 4 will be cancelled out in the equation and if it's 6, it will be too? 
Please advise.
Sorry for asking such a trivial question. 

As a matter of personal preference, I like $b$ to represent the base rather than $a$.
So then $(b + 1)^2 = b^2 + 2b + 1$. And $(b^2 + 2b + 1)^2 = b^4 + 4b^3 + 6b^2 + 4b + 1$.
These facts are true whether $b$ is an ordinary positive integer greater than 1 or a more "exotic" number, like $\sqrt{-2}$. But whether $(b + 1)^4$ gets represented as 14641 in base $b$, that's a slightly different story.
Consider for example $b = 2$. Indeed $3^4 = 2^4 + 4 \times 2^3 + 6 \times 2^2 + 4 \times 2 + 1$. But the problem is that here it turns out that $6 > b^2$ and $4 > b$. Then $4b^3$ requires more than four binary digits to represent, $6b^2$ requires more than three bits and $4b$ requires more than two bits. So the binary representation of 81 is 1010001 rather than 14641.
Some of these problems persist through $b = 6$, because although $4b^3 < b^4$, we still have $6b^3 > b^3$. Then 2401 is 15041 in base 6.
The smallest integer $b$ satisfying $4b^3 < b^4$, $6b^2 < b^3$ and $4b < b^2$ is $b = 7$. Indeed 4096 in base 7 is 14641.
