Wanted to add yet another method to the "catalogue"; I learned this by solving a problem from the book Numbers and Functions: Steps into Analysis by R P. Burn.
First, we have the following lemma:
Let $0 < a < b$. Then $\frac{b^{n+1}-a^{n+1}}{b-a} < (n+1)b^n$
This can be proved by noting that $b^{n+1}-a^{n+1} = (b-a)(b^n + b^{n-1} a + b^{n-2} a^2 + \dots + a^n)$ and since $0 < a < b \implies 0 < a^n < b^n$, we get $b^n + b^{n-1} a + b^{n-2} a^2 + \dots + a^n < b^n + b^{n-1} b + b^{n-2} b^2 + \dots + b^n = (n+1)b^n$.
Now plug in $a = 1 + \frac{1}{n+1}, \hspace{10pt} b = 1 + \frac{1}{n}$ into the lemma, and let $t_n = \left(1 + \frac{1}{n}\right)^n$ This gives
$\frac{\overbrace{\left(1+\frac{1}{n}\right)^{n+1}}^{(1+\frac{1}{n})t_n} - \overbrace{\left(1 + \frac{1}{n+1}\right)^{n+1}}^{t_{n+1}}}{\frac{1}{n}-\frac{1}{n+1}} < (n+1)\overbrace{\left(1+\frac{1}{n}\right)^n}^{t_n}$ which after some straightforward algebraic simplification yields $t_n < t_{n+1}$.