I am trying to prove this statement for all $ n \geq 1 $ using induction: $ \left( 1 + \frac{1}{n} \right)^{n} \leq \sum_{k=0}^{n} \frac{1}{k!} < 3. $
I said:
Base case $ n = 1 $: $ \left( 1 + \frac{1}{1} \right)^{1} \leq \sum_{k=0}^{1} \frac{1}{k!} < 3, $ which is okay.
Induction step: Suppose that $ \displaystyle \left( 1 + \frac{1}{n} \right)^{n} \leq \sum_{k=0}^{n} \frac{1}{k!} < 3 $ for a given $ n \in \mathbb{N} $.
Transition from $ n \to n + 1 $: $ \displaystyle \left( 1 + \frac{1}{n + 1} \right)^{n+1} = \left( 1 + \frac{1}{n + 1} \right)^{n} \left( 1 + \frac{1}{n + 1} \right) = \ldots \text{Help} \ldots < 3. $
I need some guidance for proof-writing (-thinking) in orders.