```
___________________________________
|                                   |
|   oo                              |
|  ---                              |
|  \     1                          |
|   |   ---   =   lim ( 1 + 1/n )n  |
|  /     n!      n->oo              |
|  ---                              |
|  n=0                              |
|___________________________________|
```
I did a proof (demonstration, actually, convergance isn't formally considered) of this while paying quite much too little attention in grade 11 math... I don't know whose it actually is. Probably no-one's, it's really straightforward. Phun, It's done from a high school perspective, so excuse the conceptual inelegance.

Now, since n is approaching infinity, applying the Binomial Theorem to the right side of the above equation looks promising... It'll give you a sum of an infinite number of terms. Let's take a look, see.

The Binomial Theorem:
```              n
---
\       n!
(a+b)n =      |  ----------- * an-k * bk
/    k!*(n-k)!
---
k=0
```
Let's calculate a few terms of the expansion of e, using the theorem, and starting at tn (k = n).
```tn (k = n)
___________________________________________________
|                                                   |
|                      n!                           |
|  (1/n+1)n = lim  ----------- * (1/n)n-n * (1)n     |
|            n->oo  n!*(n-n)!                       |
|___________________________________________________|
```
Clean it up, and it all comes out to 1 (1/(0!)).

tn-1 (k = n-1)
``` _____________________________________________________________
|                                                             |
|                          n!                                 |
|  (1/n+1)n = lim  ------------------- * (1/n)n-(n-1) * (1)n-1  |
|            n->oo  (n-1)!*(n-(n-1))!                         |
|_____________________________________________________________|

n(n-1)!
(1/n+1)n = lim  ------------- * (1/n)1 * (1)n-1
n->oo   (n-1)!*1!

(1/n+1)n = lim  n * (1/n) * 1 = 1
n->oo
```
...and 1 is 1/(1!)

tn-2 (k = n-2)
```______________________________________________________________
|                                                              |
|                           n!                                 |
|  (1/n+1)n = lim  ------------------- * (1/n)n-(n-2) * (1)n-2  |
|            n->oo  (n-2)!*(n-(n-2))!                          |
|______________________________________________________________|

n(n-1)(n-2)!
(1/n+1)n = lim  -------------- * (1/n)2 * (1)n-2
n->oo   (n-2)!*2!

n(n-1)     1
(1/n+1)n = lim  -------- * --- * 1
n->oo    2!        n2                                               ```
We can use the limit product law to rearrange this and divide it into two seperate limits:
```
1          n(n-1)
(1/n+1)n = lim  --- *  lim -------- * 1
n->oo  2!   n->oo   n2
```
The first limit is simply 1/2!, by the constant limit theorem. The second can easily be show to be 1 by multiplying through by 1/n and taking the limit... intuitively, think of how the ratio between n and n-1 shrinks to zero as n approaches infinity.

A similar argument can be used for all following terms
```                                                     oo
---
1     1     1     1           \     1
lim   ( 1 + 1/n )n = --- + --- + --- + --- + ... =  |   ---
n->oo                 0!    1!    2!    3!          /     n!
---
n=0
```
Whee.