smallest natural $n>1$ such that $2^n>n^{1000}$? Find the smallest natural number $n>1$ such that
$2^n > n^{1000} $ 

It seems with numeric methods that the correct answer is 13747.

But I'm still puzzled what is the correct rigorous analytical aproach to solve this particular problem and other similar problems?
 A: If you look at the problem from an algebraic point of view, basically the question is to solve $$f(x)=2^x-x^{1000}=0$$ The solution of this problem is analytical in term of Lambert function and the solution is given by $$x=-\frac{1000}{\log (2)} W_{-1}\left(-\frac{\log (2)}{1000}\right)$$ which, numerically equals $13746.8$.   
A quick check : $f(13746)=-5.923\times 10^{4137}$ and $f(13747)=2.029\times 10^{4137}$.  
I really enjoy these small numbers (the mass of the universe is close to $10^{53}$ kg) !  
So, the solution of the problem is $n=13747$.  
Added later 
For the considered branch of the Lambert function, an approximation is given by
$$W_{-1}(x)=L_1-L_2+\frac{L_2}{L_1}+\frac{(L_2-2) L_2}{2 L_1^2}+\frac{(2 L_2^2-9L_2+6) L_2}{6 L_1^3}+ ...$$  with $L_1=\log (-x)$ and $L_2=\log (-\log (-x))$; for this specific case, this approximation leads to a value of $13746.4$. 
A: One way to obtain the bounds of the least possible $n > 1$ such that $2^n > n^{1000}$, is to take $\log_2$ of both sides:
$$n\log_2{2}>1000\log_2{n}\\
n > 1000\log_2{n}$$
We let $n = 2^a$ for some positive integer $a$. Then
$$2^a > 1000a$$
Because $2^a$ grows a lot faster than $1000a$, it is a simple task to test values to find that the least value of $a$ is $14$. From this we see that the least valid $n$ is bounded by : $2^{13} < n_{\text{min}} \le 2^{14}$, or $8192 < n_{\text{min}} \le 16384$. Clearly, $13747$ conforms to this range.
As sas suggested, we can go on to improve this bound. Recall that $n > 1000\log_2{n}$, implying $\frac{n}{1000} > a$. If $n < 13000$, then $\frac{n}{1000} < 13$. Hence $n \ge 13000$. Similarly, if $n > 14000$, then $\frac{n}{1000} > 14$. From this we have a stronger bound for $n$ : $n \le 13000 \le 14000$.
Apart from calculating these bounds, I do not think that there is a way to actually compute by hand the smallest possible integer value of $n$, unless perhaps you want to do Newton-Rhapson's by hand.
A: I don't think there's any analytical method for computing the exact value, since your problem is equivalent to $\dfrac{\ln n}n<\dfrac{\ln2}{10^3}$ , but the function $f(x)=\dfrac{\ln x}x$ does not possess an inverse expressible in terms of elementary functions.
