Finding the first digit of $2015^{2015}$ Can anyone help me find the first digit of $2015^{2015}$?
It is easy to find the last digit but I have no idea for the first digit.
 A: The units digit is of course a 5. 
For the leading digit, see that $n^n = 10^{n\cdot log_{10}(n)} \approx 10^{2015\cdot3.304275} \approx 10^{6658.114} = 10^{6658}10^{0.114} \approx 1.3008\cdot 10^{6658}$. Therefore, the first digits of your number are 13008...
Check: http://www.wolframalpha.com/input/?i=2015^2015
A: Take the base-ten logarithm.
By the log laws, that will equal $2015\log2015$.  Suppose that is $x+y$, where $x$ is a whole number and $0<y<1$.
The number has $x+1$ digits.
$y$ tells you the first digit.  Or rather, $10^y$ starts with the same digits $2015^{2015}$ does.
A: Now, you didn't say "by hand" as these calculations often do, but I'll assume that at least you're from before 1950 so that solutions like just asking Wolfram Alpha about a mere 8000-digit number are beyond you.
We can calculate $\log_{10}(2015^{2015})=2015(\log_{10}(2.015)+3)$, and consulting my trusty tables I find that $$\log_{10}(2.015)\approx\frac{\log_{10}(2.01)+\log_{10}(2.02)}2\approx\frac{0.30320+0.30535}2\approx0.30428,$$ so $\log_{10}(2015^{2015})\approx6658.1$. (I need to do this calculation well enough to show it is between $6658$ and $6658+\log_{10} 2\approx6658.3$ to prove the claim.)
Then, $6658\le\log_{10}(2015^{2015})<6658+\log_{10} 2$ gives $10^{6658}\le2015^{2015}<2\cdot 10^{6658}$, so the first digit is $1$.
