discrete mathematics - How to get $k^{k + 1} + k^k$ to equate $(k+1)^{k+1}$?

This is a problem from Discrete Mathematics and its Applications

18. Let $P(n)$ be the statement that $n!
    

$\quad(a)$ What is the statement $P(2)$?
$\quad(b)$ Show that $P(2)$ is true, completing the basis step of the proof.
$\quad(c)$ What is the inductive hypothesis?
$\quad(d)$ What do you need to prove in the inductive step?
$\quad(e)$ Complete the inductive step.
$\quad(f)$ Explain why these steps show that this inequality is true whenever $n$ is an integer greater than $1$.

I am currently on part e, completing the inductive step.
Here is my work so far,

I was able to show that the basic step, $P(2)$ is true because $2! < 2^2$ or $2 < 4$
Now I am trying to show the inductive step, or $P(k)\to P(k+1)$
Assuming $P(k)$, $k! To get $(k+1)!$ on both sides, I multiplied both sides by $k +1$ to get
$$(k+1)! < k^k(k+1)$$ or $$(k+1)! < k^{k + 1} + k^k$$

How can I get this expression, $k^{k + 1} + k^k$ to equate $(k+1)^{k+1}$?

Answer

$$(n+1)!=(n+1) n! < (n+1) n^n<(n+1) (n+1)^n=(n+1)^{n+1} $$