What's the first digit of 2410^2410? The first digit means the left most digit. 2410 is just an example and it can be replaced by any other numbers. 
Can any one help me to solve it?
 A: Using a calculatior we get that $\log_{10} (2410^{2410})=8150.66107260543188\ldots$.
Thus, $2410^{2410}$ has $8151$ digits, and the first one is $4$, since $10^{0.66107260543188}=4.582184853626742\ldots$.
In general, if we have a huge number $M$, then its first digit is the same as the first digit of $10^{\log_{10} M-\lfloor\log_{10} M\rfloor}$, where $\lfloor\cdot\rfloor$ is the integer part.
Incidentally, the second digit is 5, the third 8, the fourth 2 etc...
A: I'll try without computer as exposed here (the idea is a combination of Prahlad and Lucian's suggestiond) but it's not really easy... (with $\,\log(x)=\log_{10}(x)=\dfrac{\ln(x)}{\ln(10)}$ and since $\,\dfrac 1{\ln(10)}\approx 0.43$) :
\begin{align}
2410\cdot\log(24.1)&\approx 2410\cdot(\log(3)+3\log(2)+\log\left(1+\frac1{240}\right))\\
&\approx 2410\cdot(0.47712+3\cdot 0.30103+0.43/240)\\
&\approx 2410\cdot (0.47712+0.90309+0.0018\\
&\approx 2410\cdot 1.382\\
&\approx 241\cdot 13.82\\
&\approx 241\cdot 13+241\cdot 0.82\\
&\approx \text{####}.62\\
\end{align}
The first digit is obtained by computing $10^{\text{decimal part}}\approx 4$ (since $\log(4)\approx 0.60206$ and $\log(5)\approx 0.69897\approx 1-0.30103$).
(I had some luck here since the exact computation of $2410\log(24.1)= 3330.661072\cdots$ with $10^{0.661072\cdots}= 4.58218\cdots$)
