# prove that log((n^2)!)= o(log((n!)^2))

i have a question - how i can prove that:

$$\log((n^2)!) =\theta (log((n!)^2))$$

i try something like that:

$$\log((n^2)!) = 2*(log(n)!)=\theta(2*(log(n)!)=\theta(n\ log(n))$$

$$\ \theta(log(n!)^2)=\theta(log(n!)*log(n!)) = \theta(n\ log(n))$$

then we can see that is equal --> there are $$\theta(n\ log(n))$$

this is true?

• log((𝑛2)!)=2∗(𝑙𝑜𝑔(𝑛)!) is wrong. – gnasher729 Jan 31 at 21:23
• You can't do that, because $(n!)^2 < (n^2)!$ – HEKTO Jan 31 at 21:26
• Not just <, but an awful lot less. – gnasher729 Jan 31 at 21:35
• Welcome to StackOoverflow! This question has been downvoted due to poor formatting. It is not clear if you mean "big theta of" or "little o of" – Ṃųỻịgǻňạcểơửṩ Jan 31 at 21:42

$$(n^2)!$$ is the product of the numbers 1 to $$n^2$$.
$$(n!)^2$$ is the product of the numbers 1 to n, multiplied by the product of the numbers 1 to n again.
In the first product, we have $$n^2 - n$$ numbers ≥ n plus some others. In the second product we have 2n numbers ≤ n. So if we take the logarithm, the first is ≥ $$(n^2 - n) log n$$, the second is ≤ $$2n log n$$. So f(n) / g(n) ≥ (n-1)/2. Absolutely not Big-$$\Theta$$.
Alternatively: $$(n^2)!$$ is the product of $$n^2$$ numbers. We can split the product into n groups 1 .. n, n+1 .. 2n, 2n+1 .. 3n, $$n^2-n+1$$ .. $$n^2$$. The product of the numbers in each group is ≥ n!, actually significantly greater than n! in some cases. So $$(n^2)! ≥ (n!)^n$$.
If we let f(n) = $$log ((n^2)!)$$ and g(n) = $$log((n!)^2)$$, then f(n) ≥ n log(n!), while g(n) = 2 log(n!), so f(n) / g(n) ≥ n/2. So f(n) is not o (g(n)), or $$\Theta(g(n))$$, but g(n) = o (f(n)).