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Show that $\forall k \in \mathbb{N}$, $(m + 1)^{k} \cdot (k – 1)^{k} \cdot (k + 2)^{k} > 4^{k}$

Prove that, for any $k\in\mathbb{N}$, $$(m + 1)^{k} \cdot (k – 1)^{k} \cdot (k + 2)^{k} > 4^{k}$$

An attempt using by induction:
Base case:
$$(m + 1)^{k} \cdot (k – 1)^{k} \cdot (k + 2)^{k} > 4^{k}$$
Induction step:
$$(m + 1)^{k} \cdot (k – 1)^{k} \cdot (k + 2)^{k} > 4^{k} \Rightarrow$$
$$(m + 1)^{k-1} \cdot (k – 1)^{k-1} \cdot (k + 2)^{k-1} \cdot (m + 1) \cdot (k – 1) \cdot (k + 2) > 4^{k – 1} \cdot (k + 2)^{k-1}$$
Here I stuck, I don’t know how to move on.

A:

It is useful to note that
$$
k\ge 2\Longrightarrow (k-1)^{k-1}>1
$$
and then if $m>2$ we can write
$$
(m+1)^k=\left(m^k+m^{k-1}\right)>4^k+4^{k-1}=4^k-2\cdot4^{k-1}
$$

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