Induction Proof, Problem given. 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): 

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