Answer
$x^{20}$ grows slower than $1.00001^x$.
Work Step by Step
We will find the limit $\lim_{x\to\infty}\frac{x^{20}}{1.0000^x}.$
1) If it is equal to zero then $x^{20}\ln x$ grows slower than $1.00001^x$;
2) If it is equal to $\infty$ then $x^{20}$ grows faster than $1.00001^x$;
3) If it is equal to some non zero constant then their growth rates are comparable.
"LR" will stand for "Apply L'Hopital's rule":
$$L=\lim_{x\to\infty}\frac{x^{20}}{1.00001^x}=\lim_{x\to\infty}\left(\frac{x}{1.00001^{x/20}}\right)^{20}=\left(\lim_{x\to\infty}\frac{x}{1.00001^{\frac{x}{20}}}\right)^{20}=l^{20},$$
where we have denoted $l=\lim_{x\to\infty}\frac{x}{1.00001^{\frac{x}{20}}}$. Now we have:
$$l=\lim_{x\to\infty}\frac{x}{1.00001^{\frac{x}{20}}}=\left[\frac{\infty}{1.00001^{\infty/20}}\right]=\left[\frac{\infty}{\infty}\right][\text{LR}]=\lim_{x\to\infty}\frac{(x)'}{(1.00001^{\frac{x}{20}})'}=\lim_{x\to\infty}\frac{1}{1.00001^{\frac{x}{20}}\ln1.00001(x/20)'}=\lim_{x\to\infty}\frac{1}{1.00001^{\frac{x}{20}}\ln1.00001\cdot\frac{1}{20}}=\lim_{x\to\infty}\frac{20}{1.00001^{x/20}\ln 1.00001}=\left[\frac{20}{1.00001^{\infty/20}\ln1.00001}\right]=\left[\frac{20}{\infty}\right]=0,$$
where we have used that $1.00001^x$ tends to $\infty$ when $x\to\infty$ because the base is greater then $1$. Also $\ln1.00001>0$ because the argument of the logarithm is greater than $1$. Putting this into the initial limit we get
$$L=l^{20}=0^{20}=0,$$
and thus $x^{20}$ grows slower than $1.00001^x$.
