How would you prove that $2^{50} < 3^{33}$ without directly calculating the values Could you generalise the question and get something along the lines of $n^{50} < (n+1)^{33}$ ?
 A: $ 2^{11} = 2048 < 2187 = 3^7$ implies $2^{44} < 3^{28}$.
$ 2^{6} = 64 < 243 = 3^5$ implies $2^{50} < 3^{33}$.
A: The methods given in the other answers are all very complicated.  Furthermore, as Did points out in a comment they all depend on facts which are not in principle any less complex than the statement that is to be proved.  The following method is quite simple and satisfies the request with no advanced theory whatever and “without calculating the values” as required:
Take a heap of red beans of size $2^{50}$ and a heap of navy beans of size $3^{33}$.  Repeatedly remove one bean from each pile until the red pile is exhausted.  At that point some navy beans will remain and the claim is proved.
A: The easiest way is to derive the inequality from smaller ones. First you compute
$$2^3 < 3^2 \tag 1$$
and
$$2^{11} < 3^7 \tag 2$$
then you can combine these by multiplying inequalities (1) twice and (2) once:
$$2^{17} < 3^{11} \tag 3$$
Cube this to get
$$2^{51} < 3^{33}$$
which implies the original inequality.
A: \begin{align}
3^2=2^3+1\quad&\Leftrightarrow\quad \underbrace{(3^2)^{17}}_{3^{34}}=(2^3+1)^{17}=[\text{binomial}]=2^{51}+17\cdot 2^{48}+\text{positive}\quad\Rightarrow\quad \\
&\Rightarrow\quad 3\cdot 3^{33}>2\cdot 2^{50}+\frac{17}{4}\cdot 2^{50}>3\cdot 2^{50}. 
\end{align}
A: If we are allowed to "prove" with a calculator, then we have
$$\ln 2^{50}=50\ln2\approx 34.65735903$$
and
$$\ln 3^{33}=33\ln 3\approx 36.25420553$$
As $\ln2^{50}<\ln3^{33}$, $2^{50}<3^{33}$.
For the generalization. Note that
$$3^{50}=9^{25}>8^{25}=4^{\frac{3}{2}\times 25}>4^{33}$$
So it is not true for $n=3$.
Note that $n^{50}>(n+1)^{33}$ if and only if $50\ln n>33\ln(n+1)$.
Let $f(x)=50\ln x-33\ln(x+1)$. Then
$$f'(x)=\frac{50}{x}-\frac{33}{x+1}=\frac{17x+50}{x(x+1)}$$
which is positive for all $x>0$. $f$ is strictly increasing for $x>0$.
As $3^{50}>4^{33}$, $f(3)>0$. Hence $f(x)>0$ for all $x\ge 3$.
The inequality $n^{50}<(n+1)^{33}$ does not hold for $n>2$.
A: $3^7>2^{11}$.
Thus, $3^{35}>2^{55}$ and since $2^5>3^2$, we are done!
