When I was observing the increment trend difference between exponential functions and power functions, I decided to try solving the roots of the two equations. The most apparent way is by Lambert W function, which I do something as follows:
$e^x = x^4$
$(1/x) \ln(1/x) = -1/4$
$x = e^{-W(-1/4)}$
Evaluating the answer in Wolfram Alpha, I got $1.4296$ something, which is one of the three roots observed in the graph $e^x = x^4$. How do I proceed from this point? Do I try to expand it into a Taylor series and see how it goes? That would be tedious, but is that the right way?
1.)
$$e^x=x^{nx}\ \ \ (n\in\mathbb{N}_+)$$ $$e^x=e^{nx\ln(x)}$$ $$x=e^\frac{1}{n}$$
Now I assume you are interested only in the solutions in the reals.
2.)
$$e^x=x^4$$
See the graph of the function $x\mapsto e^x-x^4$ in the reals. There are not too much extremas. Therefore you could solve the equation numerically.
There are at least two ways of solving in closed form.
a)
$$e^x=x^4$$ $$x=4\ln(|x|)$$ $$\frac{1}{x}\ln(|x|)=\color{red}+\frac{1}{4}$$ $$x=-4W_0\left(\pm\frac{1}{4}\right),-4W_{-1}\left(-\frac{1}{4}\right)$$ $\ $
$$e^{-W_{0/1}\left(\pm\frac{1}{4}\right)}=-4W_{0/-1}\left(\pm\frac{1}{4}\right)$$
b)
$$e^x=x^4$$ $$e^{\frac{1}{4}x}=|x|$$ $$1=|x|e^{-\frac{1}{4}x}$$ $$|x|e^{-\frac{1}{4}x}=1$$ $$-\frac{1}{4}|x|e^{-\frac{1}{4}x}=-\frac{1}{4}$$ $$x=-4W_0\left(\pm\frac{1}{4}\right),-4W_{-1}\left(-\frac{1}{4}\right)$$
We can represent a closed-form function (like Lambert W or the natural exponential function) by its series representations, its differential or integral equations and its algebraic transformation rules.
If you want numeric values for your closed-form solutions, you have to use the series representations of Lambert W.