Proof by induction $(1+\frac{1}{n})^n < \sum^n_{i=0} \frac{1}{i!}$, $\forall n\ge2$

Question

I'm stuck with this proof, this is what I did:

$n=2: \bigg(1+\cfrac{1}{2}\bigg)^{2} < 1 + 1 + \cfrac{1}{2}$

where verifies: $\cfrac{9}{4} < \cfrac{5}{2}$

Now with the hypothesis: $\bigg(1+\cfrac{1}{k}\bigg)^{k} < \sum_{i=0}^k \cfrac{1}{i!}$

then: $\bigg(1+\cfrac{1}{k+1}\bigg)^{k+1} < \sum_{i=0}^{k+1} \cfrac{1}{i!}$

$\bigg(1+\cfrac{1}{k+1}\bigg)^{k+1}=\bigg(1+\cfrac{1}{k+1}\bigg)^{k}\bigg(1+\cfrac{1}{k+1}\bigg)<\bigg(1+\cfrac{1}{k}\bigg)^{k}\bigg(1+\cfrac{1}{k+1}\bigg)<\sum_{i=0}^{k} \cfrac{1}{i!}\bigg(1+\cfrac{1}{k+1}\bigg)=\sum_{i=0}^{k} \cfrac{1}{i!} +\cfrac{1}{k+1}\sum_{i=0}^{k} \cfrac{1}{i!}$

and i don't know how to relate the second element and get $\cfrac{1}{(k+1)!}$. Should I use Newton binomial?

Thanks in advance.

Answer 1

Although not by induction, here is a proof,

$$(1+1/n)^n = \sum_{i=0}^n {n \choose i} \frac{1}{n^i} = \sum_{i=0}^n \frac{1}{i!} \frac{n(n-1)(n-2)...(n-i+1)}{n^i} < \sum_{i=0}^n \frac{1}{i!} $$