The number e is transcendental

In the proof we shall use the standard notation  to denote the th derivative of  with respect to .

Suppose that  is a polynomial of degree  with real coefficients. Let  . We compute ; using the fact that  (since  is of degree  ) and the basic property of , namely that , we obtain  

The mean value theorem asserts that if  is a continuously differentiable, single-valued function on the closed interval  then

We apply this to our function  which certainly satisfies all the required conditions for the mean value theorem on the closed interval  where  and  where  is any positive integer. We then obtain that  where  depends on  and is some real number between 0 and Multiplying this relation through by  yields  We write this out explicitly(1):

Suppose now that  is an algebraic number; then it satisfies some relation of the form (2)

where  are integers and where 

In the relations (1) let us multiply the first equation by , the second by , and so on; adding these up we get

In view of relation  whence the above equation simplifies to (3)

All this discussion has held for the  constructed from an arbitrary polynomial . We now see what all this implies for a very specific polynomial, one first used by Hermite, namely,

Here  can be any prime number chosen so that  and . For this polynomial we shall take a very close look at  and we shall carry out an estimate on the size of 

When expanded,  is a polynomial of the form

where  are integers.

When  we claim that  is a polynomial, with coefficients which are integers all of which are multiples of 

Thus for any integer  for  is an integer and is a multiple of 

Now, from its very definition,  has a root of multiplicity  at  Thus for  However,  by the discussion above, for  is an integer and is a multiple of 

What about  Since  has a root of multiplicity  at  For  is an integer which is a multiple of  But  and since  and is a prime number,  so that  is an integer not divisible by . Since  we conclude that  is an integer not divisible by  Because  and  and because  whereas  we can assert that  is an integer and is not divisible by .

However, by  What can we say about  Let us recall that

where  Thus

As 

whence we can find a prime number larger than both  and  and large enough to force  But  so must be an integer; since it is smaller than 1 in size our only possible conclusion is that . Consequently,  this however is sheer nonsense, since we know that , whereas  0. This contradiction, stemming from the assumption that  is algebraic, proves that  must be transcendental.