Which is Bigger?

Relevant wiki: Exponential Inequalities - Different Base

Let $11^{33} \ ? \ 2001^{10}$

or, $(11^3)^{11} \ ? \ 2001^{10}$

or, $1331^{11}\ ? \ 2001^{10}$

Now dividing both sides with $1331^{10}$ we find,

$1331 \ ? \ (2001/1331)^{10}$

Notice $(2001/1331) < 2$ so, $(2001/1331)^{10} < 2^{10}=1024$ .

So, $1331 > 1024 > (2001/1331)^{10}$ /

or $1331 > (2001/1331)^{10}$ , so now can do backward substitution & reach the following result:

$11^{33} > 2001^{10}$

$\begin{aligned} \frac {11^{33}}{2001^{10}} & = \frac {(10+1)^{33}}{(2\cdot 10^3+1)^{10}} \quad \quad \small \color{#3D99F6}{\text{By binomial expansion}} \\ & = \frac{10^{33}+33\cdot 10^{32}+528\cdot 10^{31}+...}{2^{10}10^{30}+10\cdot 2^9 10^{27}+45\cdot 10^{24}+...} \\ & = \frac{10^3 10^{30}+3300 \cdot 10^{30}+5280 \cdot 10^{30}+...}{1024 \cdot 10^{30}+5.120 \cdot 10^{30}+45\cdot 10^{24}+...} \\ & = \frac {9580\cdot 10^{30}+...}{1029.12\cdot 10^{30} +...} \\ & > 1 \\ \implies 11^{33} & \boxed{>} 2001^{10} \end{aligned}$

11^3 = 1331,
2001^10 < {(2)(11^3)}^10
= (2^10)(11^30) = (1024)(11^30).
< (1331)(11^30) = (11^3)(11^30)=11^33
So, (2001)^10 < (11)^33.