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.
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1$\begingroup$ So $(k + 1)! > (k + 1)^{2} \iff k! > k + 1 \iff (k - 1)! > 1 + \frac{1}{k}$. That should give you a start. $\endgroup$– AJYFeb 5, 2016 at 18:37
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2$\begingroup$ $k\ge 3$ works fine. $\endgroup$– lhfFeb 5, 2016 at 18:46
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3 Answers
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HINT:
Use the relationship
$$(k+1)!=(k+1)k!$$
answered Feb 5, 2016 at 18:38
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$\begingroup$ Since $k! > (k + 1)$, does this prove the statement? Or would I have to go further? $\endgroup$– 123Feb 5, 2016 at 18:59
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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...?
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Hint: For which $k$ does this argument work? $$ k! > 2k \ge k+1 $$
