# How to get $k^{k + 1} + k^k$ to equate $(k+1)^{k+1}$?

This is a problem from Discrete Mathematics and its Applications

1. Let $$P(n)$$ be the statement that $$n!, where $$n$$ is an integer greater than $$1$$.

$$\quad(a)$$ 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 $$P(k)$$, $$k! , show $$(k+1)! < (k+1)^{k+1}$$
To get $$(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 $$(k+1)^{k+1}$$?

• Should not have expanded. We have $(k^k)(k+1)\lt (k+1)^k(k+1)=(k+1)^{k+1}$. – André Nicolas Feb 25 '15 at 1:47
• You don't need to "equate", it is sufficient to establish that $k^{k+1} + k^k < (k+1)^{k+1}$ – DanielV Feb 25 '15 at 1:49
• @DanielV wait how did you turn that screenshot to text content? – committedandroider Feb 25 '15 at 1:52
• @committedandroider Mr. Scott did that (and well), and to see how he did it, click the edit button. – DanielV Feb 25 '15 at 1:53

## 2 Answers

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

• Oh cause you know (n)$^n < (n+1)^n$? for integers > 1? – committedandroider Feb 25 '15 at 1:54

You have $$k^{k+1} + k^k = k^k(k+1)$$

Can you use this fact?