Product Of Series With Increment Powers I found this interesting aptitude question and I don't know how to solve this genre of question. Any help is welcome :)
$$\prod_{n=1}^{49}n^n=1¹\cdot 2²\cdot\ldots\cdot49^{49}=?$$
Thanks.
 A: As suggested by Semiclassical, let  $$X=\prod_{n=1}^{49}n^n=1^1\cdot 2^2\cdot\ldots\cdot49^{49}$$ so $$\log(X)=\sum_{n=1}^{49} n\log(n)$$ which is much easier to handle with a pocket calculator. Being patient for a couple of minutes, you should get that $\log(X)=4167.802939$, from which $$X \approx 1.131926083941\times 10^{1810}$$ while the approximate value of the original product is $1.131925630\times 10^{1810}$.
May be, you could be interested by the fact that $$\prod_{n=1}^{m}n^n=H(m)$$ where $H$ is the hyperfactorial function.
Added later
In a comment, Semiclassical spoke about a type of Stirling expansion. According to
Mathworlds page, it write $$H(m)=A e^{-\frac{m^2}{4}} m^{\frac{m^2}{2}+\frac{m}{2}+\frac{1}{12}} \left(1+\frac{1}{720
  m^2}-\frac{1433 }{7257600 m^4}+O\left(\left(\frac{1}{m}\right)^5\right)\right)$$ where $A$ is Glaisher number $(\approx 1.282427129)$. Using the first term only, for $m=49$, we then obtain $1.131924975477377\times 10^{1810}$. Notice that the second term is only $\frac{1}{1728720}$ compared to $1$.
