proof that e is the sum of the reciprocals of factorials Ok so we know that:
e= lim n→∞(1+1/n)^n.
and we know by binomial theorem, that
lim n→∞ of $\sum_{k=0}^n {n \choose k} (1/n)^k = (1+ \frac{1}{n})$
To simplify further to $\sum_{k=0}^n \frac{1}{k!} = e$ 
we must evaluate the following limit:
n→∞${n \choose k} \frac{1}{n^k}=\frac{(n)(n-1)(n-2)...(n-k+1)}{k! n^k}$
and this is supposed to equal to $\frac{1}{k!}$
This is not clear to me algebriacally, so may some one please clear this up step by step so I can understand?
 A: We have $$e=\lim_{n\to\infty}\sum_{k=0}^n\frac{\binom{n}{k}}{n^k}=\lim_{n\to\infty}\sum_{k=0}^\infty\frac{\binom{n}{k}}{n^k}=\lim_{n\to\infty}\sum_{k=0}^\infty\frac{1}{k!},$$where with $k$ fixed$$\lim_{n\to\infty}\frac{\binom{n}{k}}{n^k}=\frac{1}{k!}\lim_{n\to\infty}\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)=\frac{1^{k-1}}{k!}=\frac{1}{k!}.$$
A: We may prove that by considering the Maclaurin series of the exponential function.$$f(x)=e^x=\displaystyle \sum_{k=0}^{\infty}\dfrac{x^k}{k!}$$
$$\therefore f(1)=e=\displaystyle \sum_{k=0}^{\infty}\dfrac{1}{k!}=1+\dfrac{1}{2!}+\dfrac{1}{3!}+\cdots$$
A: Since you know that
$$\lim_{n\rightarrow \infty}\frac{c_0 + c_1n + \cdots  + c_kn^k}{n^k} = c_k$$
for any constants $c_i$, we get
$$\lim_{n\rightarrow \infty}\frac{c_0 + c_1n + \cdots  + c_kn^k}{k!n^k} = \frac{1}{k!}\lim_{n\rightarrow \infty}\frac{c_0 + c_1n + \cdots  + c_kn^k}{n^k} = \frac{c_k}{k!}$$
Note that your numerator, when expanded, is a monic polynomial with degree $k$, which fits the above with $c_k = 1$.
