# Simplify $(k +1)! > (k + 1)^2$ to prove true for $k ≥ 4$

I am trying to prove this statement is true for $k ≥ 4$. I don't know how to work with $k + 1$ factorial, so I'm a little lost on proving this.

• So $(k + 1)! > (k + 1)^{2} \iff k! > k + 1 \iff (k - 1)! > 1 + \frac{1}{k}$. That should give you a start.
– AJY
Feb 5, 2016 at 18:37
• $k\ge 3$ works fine.
– lhf
Feb 5, 2016 at 18:46

HINT:

Use the relationship

$$(k+1)!=(k+1)k!$$

• Since $k! > (k + 1)$, does this prove the statement? Or would I have to go further?
– 123
Feb 5, 2016 at 18:59
• You would have to go further. Feb 5, 2016 at 19:03
• Yes, you need to pursue a bit further. Feb 5, 2016 at 19:17

First note that $$k! = k\times \bigg((k-1)\times ... \times 1\bigg) = k \times (k-1)!$$

Now

$$(k+1)!-(k+1)^2 = \color{red}{(k+1)}k!-\color{red}{(k+1)}^2$$

$$=\color{red}{(k+1)}\big(k!-(k+1)\big)$$

Here we have taken out $(k+1)$ as the common factor.

Now, let's look at that second factor, $\big(k!-(k+1)\big)$. $$k!-(k+1) = k!-k-1$$

$$= \bigg(\color{red}{k}\cdot (k-1)! -\color{red}{k}\bigg)-1$$

$$=\color{red}{k}\bigg((k-1)!-1\bigg)-1$$

Now it is not hard to show that $${k}\bigg((k-1)!-1\bigg)-1\geq0$$ And obviously $(k+1)>0$

So...?

• Honestly, not at all, I'm pretty confused about this. I don't really understand any of those steps.
– 123
Feb 5, 2016 at 18:55
• Ill edit the answer. Feb 5, 2016 at 18:56
• @TK-421, is it a little clearer now and is there anything still confusing you? Feb 5, 2016 at 19:07

Hint: For which $k$ does this argument work? $$k! > 2k \ge k+1$$