Induction Proof, Problem given.

Question

Explanation on difficulty: For this question, it is easy for me to show the base case but I am having difficulty with the induction hypothesis.

Attempt at a solution (work attached, obviously wrong as I'm struggling but still attached):

Answer 1

Suppose

$$(n+1)^{n-1} < n^n \tag {1}$$

for some $n$. We want to show that

$$(n+2)^n < (n+1)^{n+1} \tag{2}$$

The only way I know how to get from the RHS of $(1)$ to the $RHS$ of $(2)$ is to multiply both sides of $(1)$ by $\dfrac {(n+1)^{n+1}}{n^n}$. Then $(1)$ becomes

$$ \dfrac {(n+1)^{2n}}{n^n}< (n+1)^{n+1} \tag {3}$$

To complete the proof, it is sufficient to show that the $LHS$ of $(3)$ is greater than $(n+2)^n$. In other words

$$ (n+2)^n<\dfrac {(n+1)^{2n}}{n^n} $$

Multiplying both sides by $n^n$ yields

$$ (n^2+2n)^n < (n^2+2n+1)^n $$

Which leads to

$$ 0<1$$

Of course, when you write up the formal proof, you must start from $0<1$ and work your way up.

Answer 2

So you need to prove $(k+2)^k<(k+1)^{k+1}$. Rearrange: $$\frac{(k+2)^{k+1}}{(k+1)^{k+1}}<k+2\Rightarrow$$ $$\left(1+\frac{1}{k+1}\right)^{k+1}<e<k+2, k\ge 1.$$

Answer 3

So you know base case is true

Now you have to prove that $${(k+2)}^k\lt {(k+1)}^{k+1}$$ Writing RHS as ${((k+2)-1)}^{k+1} $ and Adding both sides $(k+1){(k+2)}^k$ and expanding RHS by binomial theorem will give $$LHS={(k+2)}^{k+1}$$$$RHS={(k+2)}^{k+1}+ ..... (some ~positive ~terms)$$ Comparing RHS and LHS we can see that clearly $RHS\gt LHS$