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

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

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. $~~\blacksquare$
