Count if you can

There is another way to look at this problem and its solution:

12^12 = (12^4)^3 = (144^2)^3

But 144^2 = (1.44 * 100)^2

Using approximation, 1.44 is close to square root 2.

So, 144^2 ~= 2 * 10^4

=> (144^2)^3 ~= (2 * 10^4) ^ 3 ~= 8 * 10^12

Thus 12^12 contains 13 digits.

The number of digits is equal to $\left\lfloor 12\ \log 12 \right\rfloor.$ We make some estimates. $\log 12 > \log 10 = 1;$ $\log 12 < \log 12\tfrac12 = \log\frac{100}{2^3} = 2 - 3\log 2 > 2 - 3\cdot\tfrac3{10} = \frac{11}{10}.$ (We used $10^3 = \log 1000 < \log 1024 = 2^{10}$ , so $3 < 10 \log 2$ .)

It follows that $12\cdot 1 < 12\log 12 < 12\cdot\frac{11}{10},$ that is $12 < 12\log 12 < 13\tfrac15.$ Therefore $12^{12}$ must have 12 or 13 digits. Since 12 was not an option, we go with 13...

for every power n of 12 it has n+1 digits

thus 12th power of 12 has 13 digits