Proof that $x^k < k^x$ So, I want to prove that $x^k$ is less than $k^x$ for any $x > k$. $x$ and $k$ are both integers.
My first approach was an induction over $k$, given that the numbers are integers. I also considered the facts that given a certain $k$, $x^k$ grows slower than $k^x$ from a certain number (the limit of the division of both functions proves it). And of course both functions are always increasing. But I don't seem to be able to pull this through.
EDIT: Of course I want to prove this for any $x$ and $k$ bigger than a certain number (I think it's $k$ = 3 and $x$ > $k$ but I'm not sure)
 A: Hint: The relevant case is $x>k>1$. Note that $\log_kx>1$ and 
\begin{align}
x^k< k^x \Longleftrightarrow & \log_k(x^k)<\log_k(k^x) \\
\Longleftrightarrow & k\log_kx<x \log_k(k) \\
\Longleftrightarrow & \frac{k}{x} <\frac{\log_k(k)}{\log_k(x)}=\frac{1}{\log_k(x)}<1 \\
\end{align}
A: Notice for any $x \ge k \ge 3$, we have
$$\frac{(x+1)^k}{k^{x+1}}\bigg/\frac{x^k}{k^x} = \frac{(1+\frac{1}{x})^k}{k} \le \frac{(1+\frac{1}{k})^k}{k} \le \frac{e}{k} < 1$$
This implies for any $x > k$,
$$\frac{x^k}{k^x} 
\le \left(\frac{e}{k}\right) \frac{(x-1)^k}{k^{x-1}}
\le \left(\frac{e}{k}\right)^2 \frac{(x-2)^k}{k^{x-2}}
\le \cdots < \left(\frac{e}{k}\right)^{x-k} \frac{k^k}{k^k} 
= \left(\frac{e}{k}\right)^{x-k}< 1$$
and hence $x^k < k^x$.
A: What I think you want to prove is that a polynomial quantity is smaller than an exponential quantity. This is intro computer science. Have a look at 
http://en.wikipedia.org/wiki/Big_O_notation
and 
Working with the ~ (tilde) notation (asymptotic analysis)
A: Graphing f(x)= x^(1/x) shows that it appears to reach a max value at about 2.7183 and decreases after that.
The derivative of f(x)=x^(1/x) is rather complicated
[x^{1/x}][1/x^2](1-ln x)
 Set it equal to zero.
The parts in brackets will never be exactly zero for x > 0.
The last factor, set equal to zero is 1-ln x = 0 which gives you x = e.
The curve has one maximum value at x = e.
After that it is a decreasing curve approaching 1 as an asymptote.
Since it is decreasing after x = e, this proves the statement
x^(1/x) < k^(1/k) for x > k > e. This proves x^k < k^x.
