Let \begin{align} t_n &= \biggl( 1 + \frac{1}{n} \biggr)^{n} \\ &= 1 + 1 + \frac{1}{2!} \biggl( 1 - \frac{1}{n} \biggr) + \frac{1}{3!} \biggl( 1 - \frac{1}{n} \biggr) \biggl( 1 - \frac{2}{n} \biggr) \\ &\qquad{}+ \cdots + \frac{1}{n!} \biggl( 1 - \frac{1}{n} \biggr) \cdots \biggl( 1 - \frac{n-1}{n} \biggr). \end{align}
If $n\ge m$, $$ t_n \geq 1+1+ \frac{1}{2!} \biggl( 1 - \frac{1}{n} \biggr) + \cdots + \frac{1}{m!} \biggl( 1 - \frac{1}{n} \biggr) \cdots \biggl( 1 - \frac{m-1}{n} \biggr) = u(n,m). $$
At first glance it looks easy and intuitive. But when I tired to prove it. I couldn't do it. First, I tried to show \begin{align} t_n \geq t_m &= 1 + 1 + \frac{1}{2!} \biggl( 1 - \frac{1}{m} \biggr) + \cdots + \frac{1}{m!} \biggl( 1 - \frac{1}{m} \biggr) \cdots \biggl( 1 - \frac{m-1}{m} \biggr) \\ &\geq 1 + 1 + \frac{1}{2!} \biggl( 1 - \frac{1}{n} \biggr) + \cdots + \frac{1}{m!} \biggl( 1 - \frac{1}{n} \biggr) \cdots \biggl( 1 - \frac{m-1}{n} \biggr), \end{align} but I think second inequality is wrong, because $c \bigl( 1 - \frac{1}{m} \bigr) \leq c \bigl( 1 - \frac{1}{n} \bigr) \lt 1$, where $c \in \mathbb{R}_{\gt 0}$ s.t. $0 \lt c \lt 1$. Since, no one has ever ask this question. I believe I must be missing something (very easy).
Another easy question is $\lim\inf_{m\to\infty}(\liminf_{n\to\infty}t_n)= \liminf_{n\to\infty}t_n$ holds because $\liminf_{n\to\infty}t_n$ is constant? What if $\liminf_{n\to\infty}t_n$ doesn’t exist. Now I don’t know what is it even mean (precisely) to say it doesn’t exist.
In your first $t_{n}$ inequality, the $RHS$ is not related to $t_{m}$; the denominators within the parentheses start at completely different indices. It is a differently intended expression, as the $u(n,m)$ indicates.
All the terms are the same from $1$ to $m$, and each term afterwards in $t_{n}$ is a small positive value. So $t_{n}$ will be clearly larger.
It is a "shorter" variation of $t_{n}$ where the terms from $1$ to $m$ match up so they can be cancelled.
I don't believe the intention at all was to prove $n \geq m \implies t_{n} \geq t_{m}.$