Which number is greater, $11^{11}$ or $9^{12}$? 
Which number is greater than $11^{11}$ or $9^{12}$?

My work so far:
$11^{11}=285311670611>9^{12}=282429536481$.
But to verify the validity of equality should be in the range of easily verifiable calculations.
 A: $$
\frac{11^{11}}{9^{11}}
=\left(\frac{2}{9}+1\right)^{11}
= \sum_{k=0}^{11} \binom{11}{k} \left(\frac{2}{9}\right)^k 
> \sum_{k=0}^5 \binom{11}{k} \left(\frac{2}{9}\right)^k = \frac{177665}{19683}
> 9
$$
A: This is a variant on Servaes's answer.  Note that $3^5=243=2\cdot11^2+1$.  Using a binomial expansion and some extremely crude upper bounds, we find
$$\begin{align}
3\cdot9^{12}
&=3^{25}\\
&=(2\cdot11^2+1)^5\\
&=32\cdot11^{10}+80\cdot11^8+80\cdot11^6+40\cdot11^4+10\cdot11^2+1\\
&\lt32\cdot11^{10}+80\cdot11^8+11^8+11^8+11^8+11^8\\
&\lt32\cdot11^{10}+121\cdot11^8\\
&=32\cdot11^{10}+11^{10}\\
&=3\cdot11^{11}
\end{align}$$
and thus $9^{12}\lt11^{11}$.
A: The goal is to prove that $(1 + 2/9)^{11} > 9$.  As the left-hand side is approximately $9.091843$ this will be a bit tricky.  
The big idea in this solution is to try to exploit the fact that $(11/9)^2 = 121/81$ is just under $3/2$, since $3/2$ will be simple to work with.
Start with the inequality $3^2 \times 29 > 2^8$, i. e. $261 > 256$.  Multiply throuhg on both sides by $3^4 2^3$ to get $3^6 \times 232 > 81 \times 2^{11}$.  
Now, we can rewrite this as 
$$ 3^{11} \times {232 \over 243} > 81 \times 2^{11}. $$
We then have $(232/243) = 1-11/243 < (1-1/243)^{11}$ (by Bernoulli's inequality, as pointed out by roby5) and so it follows that
$$ 3^{11} \times (1-1/243)^{11} > 81 \times 2^{11}. $$
At this point most of the work is done.  Divide both sides by $2^{11}$ to get
$$ 1.5^{11} \times (1-1/243)^{11} > 81 $$
and multiply both sides by $81^{11}$ to get
$$ 121.5^{11} \times (1-1/243)^{11} > 81^{12}  $$
But since both factors on the left-hand side are eleventh powers, we can rewrite this as
$$ 121^{11} > 81^{12} $$
and taking square roots of both sides gives the desired result.
A: More generally, you are looking to prove: $x^x > (x-2)^{x+1}$. This can be done by taking log both sides, and its easier. Consider $f(x) = x\ln x - (x+1)\ln (x-2)$ on $(11, \infty)$, and taking log we have: $f'(x) = \ln x + 1 - \ln(x-2) - \dfrac{x+1}{x-2}$. We have $f''(x) = \dfrac{1}{x} - \dfrac{1}{x-2}+ \dfrac{3}{(x-2)^2}= \dfrac{(x-2)^2-x(x-2) + 3}{x(x-2)^2}= \dfrac{7-2x}{x(x-2)^2} < 0\implies f'(x) > f'(\infty) = 0 \implies f(x) > f(11) $
A: Just for the heck of it, here's another approach using the fact that
$$9^3=6\times11^2+3\qquad\text{ and }\qquad 6^4<11^3-3\times11.$$
It is by no means the slickest way;
\begin{eqnarray*}
9^{12}&=& (6\times11^2+3)^4=3^4\times(2\times11^2+1)^4\\
      &=& 3^4\times(2^4\times11^8+4\times2^3\times11^6+6\times2^2\times11^4+4\times2\times11^2+1)\\
&=&6^4\times(11^8+2\times11^6+\tfrac{3}{2}\times11^4+\tfrac{1}{2}\times11^2+\tfrac{1}{16})\\
&<&(11^3-3\times11)\times(11^8+2\times11^6+\tfrac{3}{2}\times11^4+\tfrac{1}{2}\times11^2+\tfrac{1}{16})\\
&=&11^{11}+2\times11^9+\tfrac{3}{2}\times11^7+\tfrac{1}{2}\times11^5+\tfrac{1}{16}\times11^3\\
&\ &\quad\ \ \ -3\times11^9-\ 6\times11^7-\tfrac{9}{2}\times11^5-\ \tfrac{3}{2}\times11^3-\tfrac{3}{2}\times11
\end{eqnarray*}
The alignment of the last expression shows that it is smaller than $11^{11}$, as desired.
