Use induction to prove the following: $1! + 2! + .... + n! \le (n + 1)!$

Question

Use induction to prove the following: $1! + 2! + .... + n! < (n + 1)!$

Base case: $n = 1$

$1! < 2!$ true

Inductive step: Assume that $1! + 2! + .... + k! \le (k + 1)!$ is true

let $n = k + 1$

$1! + 2! + .... + (k + 1)! < (k + 2)!$

$1! + 2! + .... + k! + (k + 1)! < (k + 2)!$

$1! + 2! + .... + k! + (k + 1)! < (k + 2)!$

$(k + 1)! + (k + 1)! < (k + 2)!$

$2(k + 1)! < (k + 2)!$

$2(k + 1)(k)(k - 1).. < (k + 2)(k + 1)(k)(k - 1)....$

$2 < (k + 2)$ the right is always bigger for k are non negative integer.

Answer 1

You have all the right pieces, but it's pretty sloppy. Here's a cleaned up version.

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We will prove by induction on $n \in \mathbb N$ that:

$$ 1! + 2! + \cdots + n! < (n + 1)! \tag{$\star$} $$

Base Case: Notice that $(\star)$ holds for $n = 1$, since: $$ 1! = 1 < 2 = (1 + 1)! $$ Inductive Step: Assume that $(\star)$ holds for $n = k \geq 1$.

It remains to prove that $(\star)$ holds for $n = k + 1 \geq 2$. Indeed, observe that: \begin{align*} 1! + 2! + \cdots + (k + 1)! &= (1! + 2! + \cdots + k!) + (k + 1)! \\ &< (k + 1)! + (k + 1)! &\text{by the inductive hypothesis} \\ &= 2(k + 1)! \\ &= (0 + 2)(k + 1)! \\ &< (k + 2)(k + 1)! &\text{since } k \geq 1 > 0 \\ &= (k + 2)! \end{align*} as desired. $~~\blacksquare$