Theorem 3.31 of Baby Rudin

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 $$n \geq m \implies t_{n} \geq t_{m}.$$