How I show that $571^{61} <61^{571}$ using prime number theorem? $571$ and $61$ are primes , I want really to use some basics in number theory to show that $571^{61} <61^{571}$ without using compuation by wolfram alpha or something related to computation, I'm interested if it is possible to show that using Prime number theorem ?
 A: The following doesn't use any aspects of prime numbers (your question leaves open whether a solution using the primality of $61$ and $571$ is necessary), but suffices to show that this isn't a particularly close inequality:
\begin{align}
61^{571} & > 32^{571} \\
         & = (2^5)^{571} \\
         & = 2^{2855} \\
         & > 2^{610} \qquad \text{like, a ginormous humongous lot larger} \\
         & = (2^{10})^{61} \\
         & = 1024^{61} \\
         & > 571^{61}
\end{align}
A: This is more related to analysis, rather than prime number theorem.
You just need to show, by differentiating, that the function $f(x) = \ln(x)/x$ is decreasing for $x > e$.
It follows that $\ln(571)/571 < \ln(61)/61$ and, after exponentiating, leads to your original inequality.
A: man on the street approach:
$$61^{571}>10^{300}=1000^{100}>571^{61}$$
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You can use the fact that the prime counting functions $\pi(x)$  is approximately $\frac{x}{\log x}$. So $\frac{571}{\log 571} \approx \pi(571) >\pi(61) \approx \frac{61}{\log 61}$. Now you can substitute the two approximations with actual inequalities, with some constants added in. So, already from Chebyshev we know, 
$ \frac{571}{\log 571} > \frac{8}{9} \pi(571)$ and $\frac{8}{7} \pi(61) > \frac{61}{\log 61}$. Now if you are able the connect the middle terms into an inequality, then you have the inequality for the extreme terms. It seems possible by some counting of primes. 
