Equivalent of a product Is there an asymptotic equivalent for
$$\prod_{i=1}^k i^i$$
?
Can we find something  better than $k^{k^2}$ ?
Thank you
 A: These numbers are tabulated at http://oeis.org/A002109 where it says 
$\log a(n) = 0.5 n^2 \log n - n^2/4 + O(n \log n)$ [Charles R Greathouse IV, Jan 12 2012] - corrected by Vaclav Kotesovec, Feb 20 2015. 
and also 
$a(n) \sim A  n^{n(n+1)/2 + 1/12} / \exp(n^2/4)$, where $A = 1.2824271291006226368753425\dots$ is the Glaisher-Kinkelin constant (see A074962) - Vaclav Kotesovec, Feb 20 2015. 
A: It is possible to compute equivalent of $\displaystyle{P_n=\prod_{k=1}^nk^k}$ by elementary means. Here is how :
First compute $S_n=\ln(P_n)$ :
$$S_n=\sum_{k=2}^nk\ln(k)$$
Using the fact that $x\mapsto x\ln(x)$ is increasing on $[\exp(-1),+\infty)$, we get a first (rough) estimate :
$$\int_1^nt\ln(t)\,dt\le S_n\le\int_1^{n+1}t\ln(t)\,dt$$
which leads, after a few computations, to :

$$S_n=\frac{n^2}2\ln(n)-\frac{n^2}4+o(n^2)$$

Now the goal is to refine this expansion until precision $o(1)$ is reached, so that we shall take the exponential and get an equivalent for $P_n$.
Let $$x_n=S_n-\frac{n^2}2\ln(n)+\frac{n^2}4$$
We have
$$x_n-x_{n-1}=n\ln(n)-\frac{n^2}2\ln(n)+\frac{n^2}4+\frac{(n-1)^2}2\ln(n-1)-\frac{(n-1)^2}4\sim\frac12\ln(n)$$
Hence, by partial summation of positive divergent series : $x_n\sim \frac n2\ln(n)$, so that

$$S_n=\frac{n^2}2\ln(n)-\frac{n^2}4+\frac n2\ln(n)+o(n\ln(n))$$

Next let :
$$y_n=S_n-\frac{n^2}2\ln(n)+\frac{n^2}4-\frac n2\ln(n)$$
We have :
$$y_n-y_{n-1}\sim\frac1{12n}$$
Hence, again by partial summation of positive divergent series :
$$y_n\sim\frac{1}{12}\ln(n)$$
so that :

$$S_n=\frac{n^2}2\ln(n)-\frac{n^2}4+\frac n2\ln(n)+\frac{1}{12}\ln(n)+o(\ln(n))$$

Next, let :
$$z_n=S_n-\frac{n^2}2\ln(n)+\frac{n^2}4-\frac n2\ln(n)-\frac{1}{12}\ln(n)$$
We have :
$$z_n-z_{n+1}\sim\frac1{360n^3}$$
which proves the convergence of the series $\sum z_n$ and hence the convergence of the sequence $(z_n)$ to some limit $L\in\mathbb{R}$.
So we get :

$$S_n=\frac{n^2}2\ln(n)-\frac{n^2}4+\frac n2\ln(n)+\frac{1}{12}\ln(n)+L+o(1)$$

It's time to come back to $P_n$ :
$$\boxed{P_n=\lambda e^{-n^2/4}n^{\frac{n(n+1)}{2}+\frac1{12}}}$$
where $\lambda=e^L$
