How prove this inequality?? 
How prove :


$$
{31}^{11}<{17}^{14}
$$


help please! 

 A: $$31^{11} < 32^{11} = (2^5)^{11}=2^{55} < 2^{56}=(2^4)^{14}=16^{14} < 17^{14}$$
A: $31^{11} < 34^{11}$
$34^{11} = (17\times2)^{11} = (17^{11}) \times (2^{11}) = 17^{11} \times 2^4 \times 2^4 \times 2^3$
$17^{11} \times 2^4 \times 2^4 \times 2^3 < 17^{11} \times 17 \times 17 \times 17$
$17^{11} \times 17 \times 17 \times 17 = 17^{14}$
Therefore, $31^{11} < 17^{14}$
This is easy and clear. I think this will help you.
A: Consider the :
$$1 < \left(\frac{17}{31}\right)^{11}\cdot17^{3}$$
Now the major fact:
$17^{3}>2^{11}$, i hope you could continue it!
A: $17^{14}=168377826559400929$ and $31^{11}=25408476896404831$.  QED.
A: I showed the numbers in diferent way:
${17}^{14}>{16}^{14}$
${16}^{14}={2^4}^{14}=2^{56}$
$17^{14}>2^{56}$
${31}^{11}<{32}^{11}$
$32^{11}={2^5}^{11}=2^{55}$
$2^{55}<2^{56}$
$17^{14}>2^{56}>2^{55}>31^{11}$
I think i show it more clear :)
A: $31^{11} < 32^{11} = 2^{55} < 2^{56} = 2^{4 \cdot 14} = 16^{14} < 17^{14}$
A: Consider the following equivalent inequalities
$$
\begin{align}
31^{11} &< 17^{14} \\
31^{10} \; 31 &< 17^{10} \; 17^4 \\
31 \; \sqrt[10]{31} &< 17 \; \sqrt[10]{17^4}
\end{align}
$$
and observe that the last one holds because
$$
\frac{31}{17} < 2 < \sqrt[10]{\frac{17^3}{2}} < \sqrt[10]{\frac{17^4}{31}}
$$
since $17^3/2 > 17^2 \; 8$ and $17^2 > 12^2 = 144 > 128 = 2^7$.
