Prove that $\lim \left(1 +\sum_{k = 1}^n\frac{(-1)^k}{k!}\right) = \frac{1}{e}$
As far as I know, $\lim \left(1 - \frac{1}{n} \right)^n = e$, so at that point I thought that $$\left(1 +\sum_{k = 1}^n\frac{(-1)^k}{k!}\right) = \left(1 - \frac{1}{n} \right)^n$$
I tried to use the binomial theorem here with $a = 1$ and $b = -\frac{1}{n}$.
I got $\binom{n}{0}\cdot 1 + \binom{n}{1} \cdot 1\cdot (-1)+\dotso + \binom{n}{n}(-1)^n = 1 + \binom{n}{1} \cdot 1\cdot (-1)+\dotso + \binom{n}{n}(-1)^n$. And after some time I realized my initial assumption may be wrong, because I really don't know how to proceed.
$$1+\sum_{k=1}^n\dfrac{(-1)^k}{k!}=\sum_{k=0}^n\dfrac{(-1)^k}{k!}$$
Now $$e^y=\sum_{k=0}^\infty\dfrac{y^k}{k!}$$