Monotone of a sequence without Bernouli's Inequality Can someone help prove that the sequence  $x_n=(1+\frac{1}{n})^n$ is increasing using just the binomial theorem?
I can prove that the sequence is increasing with Bernouli's inequality,but i ave a difficulty proving it with the binomial theorem.
Any help would be mush appreciated.
And also any other solution without Bernouli would be also appreciated.
Thank you in advance.
 A: Using binomial theorem, 
\begin{gather}
  x_n=\left(1+ \dfrac  1 n\right)^n=1+n\cdot  \dfrac  1
  n+ \dfrac {n(n-1)}{2!}\cdot  \dfrac  1{n^2}+ \dfrac {n(n-1)(n-2)}{3!}\cdot  \dfrac 
  1{n^3}+\ldots + \nonumber \\
  + \dfrac {n(n-1)(n-k+1)}{k!}\cdot  \dfrac  1{n^k}+\ldots + \dfrac {n(n-1)\ldots 1}{n!}\cdot  \dfrac  1{n^n}= \nonumber  \\
  =1+1+ \dfrac  1{2!}\left(1- \dfrac  1 n\right)+ \dfrac  1{3!}\left(1- \dfrac  1
  n\right) \left(1- \dfrac  2 n\right)+\ldots + \nonumber  \\
        + \dfrac{1}{k!}\left(1- \dfrac  1 n\right) \left(1- \dfrac  2 n\right)  \ldots
   \left(1- \dfrac {k-1} n\right)+\ldots + \dfrac  1{n!}\left(1- \dfrac  1
  n\right) \ldots   \left(1- \dfrac {n-1} n\right).  \label{eq_2.9}
\end{gather}
Since
$$1- \dfrac  i n<1- \dfrac  i{n+1}\quad \forall i\colon \  
 1 \leqslant i \leqslant n, $$
thus
    $$x_n<x_{n+1},\quad n=1,\,2,\,\ldots\, $$
A: The idea of solution. Let's look at $\bigl(1+\frac{1}{n-1}\bigr)^{n-1}<\left(1+\frac{1}{n}\right)^{n}$ as $1\cdot \bigl(1+\frac{1}{n-1}\bigr)^{n-1}<\left(1+\frac{1}{n}\right)^{n}$. Then we get two products of $n$ numbers with sum $n+1$. And we know that the largest one is product with all numbers are equal.
So, the solution. Using AM-GM we have
$$
\root n \of {1\cdot\underbrace{\left(1+\frac{1}{n-1}\right)\cdot \left(1+\frac{1}{n-1}\right) \cdots \left(1+\frac{1}{n-1}\right)}_{n-1}}<\frac{1+(n-1)\cdot \left(1+\frac{1}{n-1}\right)}{n}=1+\frac{1}{n},
$$
and $\bigl(1+\frac{1}{n-1}\bigr)^{n-1}<\left(1+\frac{1}{n}\right)^{n}$ 
