Show that $\left( \frac {11} {10}\right) ^{n}$ is divergent. 
Show that $\left( \dfrac {11} {10}\right) ^{n}$ is divergent.

My proof. Let $B\in\mathbb{R}$. By the Archimedean property there is a $N$ in $\mathbb{N}$ such that $N>B$.
   Let $\varepsilon >0$ By the Bernoulli inequality, we have $\left( 1+\varepsilon \right) ^{n}\geq 1+n\varepsilon$ for all $n\in\mathbb{N}$. Now, take $\varepsilon=( \dfrac {11} {10}-1)$. Then, we obtain, $\left( \dfrac {11} {10}\right) ^{n}\geq \dfrac {n} {10}+1$. So, for all $n\geq N$ we have $\dfrac {n} {10}+1>\dfrac {n} {10}>n>N>B.$ Thus, since $\left( \dfrac {11} {10}\right) ^{n}\geq \dfrac {n} {10}+1$,  $\left( \dfrac {11} {10}\right) ^{n}>B$
for all $n\geq N$.
We are done.
Can you check my proof?

Easy to think solution:
Note that $\ln$ is increasing function.
Note that $\ln\Big(\dfrac{11}{10}\Big)=\ln11-\ln10=c>0$
Now $\ln\Big(\dfrac{11}{10}\Big)^n=n(\ln11-\ln10)=nc$
Now since $c>0$, for every $N\in \mathbb{N}$ and $N>\Big\lfloor\dfrac{1}{c}\Big\rfloor+1$, you can find a $n\in\mathbb{N}$ such that $nc>N$. Hence $\ln\Big(\dfrac{11}{10}\Big)^n$ diverges to infinity. Since $\ln$ is increasing function $\Big(\dfrac{11}{10}\Big)^n$ also diverges to infinity.
