Number of digits of the number of digits of the number of digits of $2014^{2014}$ How would you solve that problem : 
What is the number of digits of the number of digits of the number of digits of $2014^{2014}$ ?
(for instance the number of digits of $12345678901234567890$ is $20$, and the numbers of digits of $20$ is $2$, and the numbers of digits of $2$ is $1$, so the number of digits of the number of digits of the number of digits of $12345678901234567890$ is $1$)
 A: We know that the number of digits in base $b$ of a given integer $N$ is $\lfloor\log_b(N) + 1\rfloor$. We also know that $\log_b(n^x) = x \log_b(n)$. Here, $n^x = 2014^{2014}$ and $b = 10$. So we can turn the ridiculously large number $2014^{2014}$ into something more manageable using logs of base $10$. That will give you the number of digits in $2014^{2014}$; the rest should follow.
Update:
To get the length of $2014^{2014}$ in base $10$, we need $\log_{10}(2014^{2014})$. Well, that is equal to $2014\cdot\underbrace{\log_{10}(2014)}$. That second term is easily handled on a computer or calculator. Now multiply the underbraced term by $2014$, add $1$, and take the integer part of that final number; that is the number of digits in $2014^{2014}$. Apply your "number of digits" reduction twice more, and you have your answer.
Update 2:
OK, you cannot use a calculator, right? but you do know how many digits are in $2014$ itself, 4, of course. That means that $3 \leq \log_{10}(2014) < 4$. Now $2014\cdot 3 = 6042$ and $2014\cdot 4 = 8056$, so no matter which way you slice it, the "number of digits of the number of digits" of $2014^{2014}$ is 4. Now you have one more reduction, I believe, right?
A: Hint:  write it as number of digits(number of digits(number of digits($2014^{2014}$)))  As usual evaluate parentheses first, so start from the inner set.  Given your comment the number of digits of $10^n$ being $n+1$, which is correct, you need to estimate number of digits($2014^{2014}$) first.
A: If you wish to determine the number of digits a given integer has in base ten, then you would use a logarithm. $$ \# \text{ of digits of } n = \lfloor  \log_{10}(n) \rfloor + 1$$
Here $\lfloor x\rfloor  $ is the largest integer smaller than $x$. For instance $5$ has one digit, and $0< \log_{10}(5) < 1$ so $\lfloor\log_{10}(5)\rfloor + 1= 1$.
You can iteratively repeat this process to find the answer you are looking for.
A: We know the answer is at least $1$ (You can't have a number with zero digits. If we define zero as having zero digits, then that would require $2014^{2014} = 0$).
Suppose the answer is at least $2$. Then the number of digits of the number of digits of $2014^{2014}$ is at least $10$. So the number of digits of $2014^{2014}$ is at least $1,000,000,000$. But, $2014^{2014} < 10000^{2014}$, the latter of which has only $8053$ digits, a contradiction.
Thus the answer is $1$.
