# How does $Θ(\log(n!))=Θ(\log(n^n)$?

How does $$Θ(\log(n!))=Θ(\log(n^n)$$?

I understand why $$Θ(\log(n!))=Θ(n\log(n))$$ and $$Θ(\log(n^n))=Θ(n\log(n))$$, therefore $$Θ(\log(n!))=Θ(\log(n^n)$$. But I am having trouble reconciling this with the following:

$$\lim_{x\to \infty}\left(\frac{\log_2(x^x)}{\log_2(x!)}\right)=\infty.$$

So how could there be constants such that $$Θ(\log(n!))=Θ(\log(n^n))$$?

Where has my understanding gone wrong?

As a matter of fact, $$\lim_{x\to \infty}\frac{\log_2(x^x)}{\log_2(x!)}=1.$$ So there is no problem to reconcile.
Looking at the first revision of the question, it seems to me that you are confused about the fact that $$\frac{n^n}{n!}$$ has different behaviour from $$\frac{\log n^n}{\log n!}$$: $$\lim_{n\to+\infty}\frac{n^n}{n!}=+\infty,\qquad\text{while}\qquad\lim_{n\to+\infty}\frac{\log(n^n)}{\log(n!)}=1<+\infty.$$ There is no problem with this. Unboundedness of ratios of functions is not preserved by applying arbitrary functions such as $$\log$$, because $$\exp$$ (the inverse of $$\log$$) is not polynomially bounded. For a simpler example, $$\lim_{n\to+\infty}\frac{n^2}{n}=+\infty,\qquad\text{while}\qquad\lim_{n\to+\infty}\frac{\log(n^2)}{\log(n)}=2<+\infty.$$
In general, a function $$h\colon\mathbb R_{>0}\to\mathbb R_{>0}$$ satifies the condition that $$\limsup_{n\to+\infty}\frac{f(n)}{g(n)}<+\infty\implies\limsup_{n\to+\infty}\frac{h(f(n))}{h(g(n))}<+\infty$$ for all $$f,g\colon\mathbb N\to\mathbb R_{>0}$$ if and only if there exists a constant $$c$$ such that $$y\le2x\implies h(y)\le ch(x)\tag1$$ for all $$x,y\in\mathbb R_{>0}$$. For example, (1) holds for all function of the form $$h(x)=x^\alpha$$ for some $$\alpha$$. On the other hand, any function $$h$$ satisfying (1) is polynomially bounded, specifically, $$h(x)=O(x^{\lceil\log_2c\rceil})$$.
$$\log (n!)=\log\left(\prod_{i=1}^{n}i\right)$$ $$=\log(1)+\log(2)+\dots+\log n$$ $$=\sum_{i=1}^{n}\log i\leq n\log n.$$ $$\sum_{i=1}^{n}\log i=\sum_{i=1}^{\frac{n}{2}}\log i+\sum_{i=\frac{n}{2}+1}^{n}\log i$$ $$\geq\frac{n}{2}\log(\frac{n}{2})=\Theta(n\log n).$$
Since $$\log n^n=n\log n$$ $$\implies \log (n!)=\Theta(\log n^n).$$