An intuitive way we can see that $(1+1/n)^n$ is increasing It can be proved that $(1+1/n)^n$ is increasing $n\in \mathbb{N}$.
Now look at the picture given below, the dark $>$ signs are actually "greater" sign, and you can check that those inequality holds perfectly!
Now take a look at at the rectangle(there those arrows are vectors). Let us define vectors on $\vec{AB}$, if the value in $A$ is greater than the value in $B$. We denote $A_{ij} =(1+1/i)^j$. $"+" \text{and}\space "="$ defined as follows: $\vec{AB}+\vec{BC}=\vec{AC}$ is value of $A>$ value in $B>$ value in $C$, implies value of $A>$ value in $C$.
Consequently we also have $A_{nn}<A_{(n+1)(n+1)}$. This completes the whole thing.(ignore the word "Consequently")
On the other way we can also say that to preserve the system $A_{nn}<A_{(n+1)(n+1)}$ has to happen.
Which one is true??

 A: A way of understanding intuitively the fact that the sequence is increasing is to note that it corresponds to componding (the same yearly) interest with greater and greater frequency.  Compounding each week will yield higher profit than compounding each month, compounding each day will yield higher profit than compounding each week, etc. A nice "formal" proof appears here.
A: Here is an easy way to see that your argument is flawed:
You could use the same argument to 'proof' that $\left(1+\frac{1}{n^2}\right)^n$ is increasing, but it is not.
A: Your diagram of inequalities is inconclusive: for each square you only have proved \begin{matrix}A_{n,n+1}&<&A_{n,n}&<&A_{n+1,n}&\text{up-left corner}\\A_{n,n+1}&<&A_{n+1,n+1}&<&A_{n+1,n}&\text{down-right corner}\\ A_{n,n+1}&&&<&A_{n+1,n}&\text{diagonal SW-NE (redundant info, btw)} \end{matrix}
You are lacking the crucial information: the diagonal NW-SE. The "Consequently" is therefore not "consequently" at all. Basically, what you made is a diagram containing all the easy eastimates hat were inconclusive in the initial attempts of solution and, not surprisingly, they still don't work.
A: You have five arrows in each square, and you seem to be saying that there is only one way to draw the sixth one. That is wrong. Consider the following two diagrams:
$$\begin{matrix}
2 & 1 \\ 4 & 3
\end{matrix} 
\qquad \text{and} \qquad
\begin{matrix}
3 & 1 \\ 4 & 2
\end{matrix} 
.
$$
The five arrows that you have in your squares all go in the same direction in the two examples. In the first example, the sixth arrow goes in the direction you would like. But in the second example, it goes in the opposite direction to the one you would like.
