Theorem 1   The sequence $(1 + 1/n)^n$ is increasing.

Proof. To see this, rewrite $1 + (1/n) = (1 + n)/n$ and divide two consecutive terms of the sequence: \begin{eqnarray*} { \left( 1 + {1 \over n} \right)^n \over \left( 1 + {1 \over n-1} \right)^{n-1} } &=& { \left( {n+1 \over n} \right)^n \over \left( {n \over n-1} \right)^{n-1} } \\ &=& \left( {(n-1)(n+1) \over n^2} \right)^{n-1} {n+1 \over n} \\ &=& \left( 1 - {1 \over n^2} \right)^{n-1} \left( 1 + {1 \over n} \right) \\ \end{eqnarray*} Since $(1 - x)^n \ge 1 - nx$ , we have \begin{eqnarray*} { \left( 1 + {1 \over n} \right)^n \over \left( 1 + {1 \over n-1} \right)^{n-1} } &\ge& \left( 1 - {n-1 \over n^2} \right) \left( 1 + {1 \over n} \right) \\ &=& 1 + {1 \over n^3}\\ &>& 1, \end{eqnarray*}hence the sequence is increasing.

Theorem 2   The sequence $(1 + 1/n)^{n+1}$ is decreasing.

Proof. By an inequality for differences of powers, we have$$ (n^2)^n - (n^2 - 1)^n \ge n (n^2 - 1)^{n-1}.$$ From this we may conclude
The last line follows from the factorization $n^2 - 1 = (n + 1) (n - 1)$ . Dividing through,$$ {n^n \over (n - 1)^n} > {(n + 1)^{n + 1} \over n^{n + 1}} ,$$ which simplifies to$$ \left( {n \over n - 1} \right)^n > \left( {n + 1 \over n} \right)^{n + 1},$$ or$$ \left( 1 + {1 \over n - 1} \right)^n > \left( 1 + {1 \over n} \right)^{n + 1}.$$