Prove that $1+{1\over1}\left(1+{1\over2}\left(1+{1\over3}\left(1+{1\over4}\left(1+{1\over5}\left( ... \right)\right)\right)\right)\right)=e$ I've found this formula, but I don't know how prove it?
What's your idea for proof?
$$1+{1\over1}\left(1+{1\over2}\left(1+{1\over3}\left(1+{1\over4}\left(1+{1\over5}\left( ... \right)\right)\right)\right)\right)=e.$$
 A: The expression in the OP is an added fraction or continued numerator, which here can be treated with a base which incresing denominator as one goes on.  It's the sort of fraction that evolved into decimals.  
You get a similar idea when you divide a length in yards into a feet, and then into b inches, then into c sixteenths, as $y \frac a3 \frac b{12} \frac {c}{16}$ yards.
A number in a reglare base B, like 10, might be written as $1 \frac aB \frac bB \frac cB \cdots $, which is a sum of fractions, the denominator of the last being the product of it and all of the ones to the right.  So in the base expression, the $c$ is divided by $B^3$.  
The expression for $e$ can be written like this:
$$ e = 1 \frac 11 \frac 12 \frac 13 \frac 14 \frac 15 \frac 16 \dots $$
In the expression for $e$, we see that it is a sum $\sum \frac 1{n!} $, which means that this expression, because that's what the running product of denominators is.
One can see from this representation, that any digit as a numerator in any place leads to a unit fraction, and thus one can write any $a/b$ in $m-1$ unit fractions, where $b\mid m!$, however, smaller solutions are allowable. (ie it's an upper limit).
