Let $S_n=\{(1+\frac{1}{n})^n\}$ and $T_n=\{1+\frac{1}{1!}+\frac{1}{2!}+\cdots+\frac{1}{n!}\}$
I am trying to prove that $\lim_{n\rightarrow \infty} S_n > \lim_{n\rightarrow \infty} T_n$
Now I know that I should use the binomial theorem expansion to compare each term from both sequences.
I know I will get $\sum_{k=0}^n \frac{n!}{(n-k)!k!}\times (\frac{1}{n})^k $ and the first few terms of the series will be $1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots$
so in my mind this looks identical to $T_n$ but I am assuming there is something that occurs later in the series that will allow me to conclude that the limit of $S_n > $ limit of $T_n$
So If you could in detail show me what that 'something' is and how you conclude that I would greatly appreciate it.
Note. In your title you said $S_n<T_n$, in the question you said $S_n>T_n$. The former is correct, and that is what I'll prove.
By the binomial theorem, $$\eqalign{\Bigl(1+\frac1n\Bigr)^n &=1+\binom n1\frac1n+\binom n2\frac1{n^2}+\binom n3\frac1{n^3}+\cdots +\binom nn\frac1{n^n}\cr &=1+\frac1{1!}\frac nn+\frac1{2!}\frac{n(n-1)}{n^2} +\frac1{3!}\frac{n(n-1)(n-2)}{n^3}+\cdots +\frac1{n!}\frac{n(n-1)\cdots1}{n^n}\cr &<1+\frac1{1!}+\frac1{2!}+\frac1{3!}+\cdots+\frac1{n!}\ .\cr}$$ Thus $S_n<T_n$ for all $n$. Note however that this does not mean $\lim S_n<\lim T_n$, it only means $\lim S_n\le\lim T_n$. And in fact the two limits are equal.
Comment. Note that $T_n$ is a finite sum, it ends with $\frac1{n!}$. From the middle of your question it looks a bit as if you think $T_n$ goes on for ever.