Question

gr8 equation proove it

Answer 1

\[\huge \prod_{n=1}^\infty e^n = \frac{1}{e^{1/12}}\]

Answer 2

\[\prod_{n=1}^\infty e^n\\
=e\cdot \prod_{n=2}^\infty e^n\\
=e\cdot e^2\cdot \prod_{n=3}^\infty e^n\\
=e^{1+2}\prod_{n=2}^\infty e^n\\
=e^{1+2+\dots}\\
= e^{\displaystyle\sum_{n=1}^\infty n}\\
=e^{-1/12}\]

Answer 3

zat it?

Answer 4

beutufel. medal+fan

Answer 5

*admires brain* @UnkleRhaukus

Answer 6

(a little bit neater)
\[\prod_{n=1}^\infty e^n\\
=e\cdot \prod_{n=2}^\infty e^n\\
=e\cdot e^2\cdot \prod_{n=3}^\infty e^n\\
=e^{1+2+3}\cdot\prod_{n=4}^\infty e^n\\
=e^{1+2+3+4+\cdots}\\
= e^{\sum_{n=1}^\infty n}\\
=e^{-1/12}\]

Answer 7

Rofl, that seems so simple in hindsight

Answer 8

the -1/12 gave it away

Answer 9

Bravo

Answer 10

We have shown that
1 > (2.718...) x (2.718... x 2.718...) x (2.718... x 2.718... x 2.718...) x ...

Answer 11

i see no prblems here B)

Answer 12

while we're at it:

\[i=e^{(\pi/2)*i}\]
therefore you can get this complex number by only raising a real number to an exponent infinitely...! :D

\[i =e^{(\pi/2)e^{(\pi/2)e^{(\pi/2)e^{\cdots}}}}\]