Cannot find a calculator that can handle 220! x 40000 For a game I am playing, I am trying to calculate the cost of purchasing all the expansion area squares on a level. the first square costs 40,000, each next square costs an additional 40,000 more than the previous square. the level is 16 x 16 squares and I start with 6 x 6 already open. I figure the math to solve is 220! x 40000, I cannot find a calulator online or irl that can handle the problem. I found a factorial calc which solves 220! but it wont let me do x 40000. 
Is there a math trick to solving this problem? or an easy way to solve it?
 A: It's $(1\color{blue}{+}2\color{blue}{+}3\color{blue}{+}...\color{blue}{+}220)×40,000$, not $(1×2×3×...×220)×40,000$.  The correct formulation should be easy to do on a calculator, especially if you recall the triangular of $n$ is $1+2+3+...+n=n(n+1)/2$.
A: You can use a trick. Use Stirling's formula:
$$n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$
You can work out the answer using only a regular calculator or even by hand.
$40000(220!) \approx 40000\sqrt{440 \pi}\left(\frac{220}{e}\right)^{220}$
$=40000\sqrt{440 \pi}10^{220\log_{10}\left(\frac{220}{e}\right)}$
$=1487172.22544 \times 10^{419.788203762}$
$=(1487172.22544 \times 10^{0.78820376210}) \times 10^{419}$
$=9131981.6054\times 10^{419}$
$\approx 9.13\times 10^{425}$
It gives a result correct to about 2 decimal places. Is that good enough?

You can do it in your head like this: $e\approx 2.7$ we find $220/2.7\approx 80$. We have $\log_{10}(80)=1+\log_{10}(8)$. We know $2^{10}=1024\approx 1000 = 10^3$. So $2\approx 10^{3/10}$ So $8=2^3\approx 2^{9/10}$. So $\log_{10}(8)\approx \frac{9}{10}$. Hence $\log_{10}(80)\approx \frac{19}{10}$. Then $220\times \frac{19}{10} = 22\times 19 = 418$ (This is one less than the correct answer for the exponent. The rest is straight forward.) 
