I'm assuming [tex]p\in\mathbb R[/tex] is fixed. First note that [tex]x^p=p^x[/tex] when [tex]x=p[/tex].
We can rewrite each expression as
[tex]x^p=e^{p\ln x}[/tex]
[tex]p^x=e^{x\ln p}[/tex]
Now to compare we only need to consider the exponents. It's easy to show that [tex]x>\ln x[/tex] for all [tex]x>0[/tex], so it follows that [tex]x^p<p^x[/tex] so long as [tex]x>p[/tex].
