*e*
is transcendental:

Let *f(x)* be a polynomial of degree *m* with reals
coefficients. Take :

By doing integration by parts m times we then get :

*(1)*

Now, let *f** be the polynomial *f *where
the coefficients have been replaced by their absolute value.

By maximising the terms in the integral we get :

*(2)*

Now all the preliminary work has been done,
lets start the battle and suppose that e is algebraic. In other terms,
suppose that there exist an integer *n>0* and non-zero *q*_{1},...,q_{n}
such that :

*q*_{0}+q_{1}e+...+q_{n}e^{n}=0*(3)*

The reste consiste of making *J=q*_{0}I(0)+q_{1}I(1)+...+q_{n}I(n)

*I(t)* does not change by definition and we choose *f(x)=x*^{p-1}(x-1)
^{p}...(x-n) ^{p} with *p* a big
prime integer.

By calculating *J* using the definition *(1)*
and the hypothesis *(3)*, we get :

where *m* is therefore the degree of
the polynomial *f* i.e. *(n+1)p-1=m*.

The *k* variables between *1* and *n* being roots of
order *p* each, we have *f*^{(j)}(k)=0 if *j<p*
and *k>0* and the same for *k=0* if *j<p-1*.
Therefore for *j#p-1* and *k#0*, either it's zero, or the
derivative take out a *p!* and so *f *^{(j)}(k)
is an integer divisible by *p!*. Further more, for the case *j=p-1*,
by recurrence on *n*, we easily get :

*f*^{ (p-1)}(0)=(p-1)!(-1)^{np}(n!)^{
p}

So, if we have *p>n*, that is *p*
does not divide *(n!)*, *f*^{(p-1)}(0)
is an integer divisible by *(p-1)!* but not by *(p!)*.

And if we then take *p>q*_{0} , *J* is a non zero integer wich is
divisible by *(p-1)!* hence, obviously, *J(p-1)!*

Here is the first inequality. Now all that remains is to find a
contradiction.

For this, since *k-n(2n)* and *m=(n+1)p-1*, by maximising
in *f**, we get *f*(k)(2n)*^{m}
and so if we use *(2)* and the definition of *J*,

*Jq*_{1}ef*(1)+...+q_{n}ne^{n}f*(n)c ^{p}

since *p>n*, *c*
being a constant independante of *p*. If we choose a large enough
*p* the facctorial being more powerful than the power, the
two inequality on *J* contradict each
other.

And there we go, end of poof.

Obviously, we get a bit the feeling that it would work for any
constant, but in truth, it's *(1)* and *(2)* that decides
everything !

is
transcendental

Always by the absurd, suppose now that is algebraic and so *=i* is also.

Let *d* be the degree of the polynomial where is the solution. Since *C* is algebraic
closed, this polynomial has d roots and denote _{1}*=, *_{2},...,_{d} all it's roots.

Then consider the minimal polynomial of , that is the smallest polynomial (non
reductable) where is a root and whose
coefficient are corprime, denote *L* it's dominant
coefficientson, that is the term of the highest power.

Knowing, as our good old Euler proved, that *exp(i)+1=0*, we can write :

*(1+exp(i*_{1}))(1+exp(i_{2}))...(1+exp(i_{d}))=0

If we expand this last expression, we get
the sum of *2*^{d}
termes *e*^{x}, where *x* is a set of values :

*x=a*_{11}+a_{22}+...+a_{dd}*a*_{i}=0
ou 1

Suppose that *n* of those values *x*
are null and denote them *ß*_{1},...,ß_{n}.

hence the sum of the *2*^{d} termes is written :

*q+exp(ß*_{1})+...+exp(ß_{n})=0*(4)*

with *q=2*^{d}-n.

As we did for *e*, the method is now to find two contradicting
inequalities for *J=I(ß*_{1})+...+I(ß_{n}),
but this time with *f(x)=L*^{np}x^{p-1}(x-ß_{1})^{p}...(x-ß_{n})
^{p}, *p* still representing a large prime
number.

By using *(1)* and *(4)* in the definition of *J*,
we get :

with as always *m=(n+1)p-1*.

If we take a closer look at the sum with *k* as an indices,we can
see that, all the roots are playing the same role as in *f(x)*,
this sum is a symetric polynomial in *Lß*_{1},...,Lß_{n},
that is it is invarient by permutation of those numbers.

But the theorem of algebra on symetric polynomials tell us that this
kind of polynomial can be witten under the form of a pollynomial with
coeeficien of the equation where *Lß*_{1},...,Lß_{n}
are roots. Hence this polynomial of coefficient is an integer, whose
sum is on *k* as well.

Then the same reasoning as for *e* is applied. We have *f *^{(j)}(ß_{k})=0
when *j<p*,
hence this sum of integers is divisible by *p!*.
By calculating from the expression of *f*, we notice that it is
also the case for the rational *f *^{(j)}(0)
if *j#p-1*. If *j=p-1*, we get :

*f *^{(p-1)}(0)=(p-1)!(-L)^{np}(ß_{1}...ß_{n})
^{p}

which is divisible by *(p-1)!* but not
by *p!* for large enough *p*.

And so for *p>q*, we have *J(p-1)!*

But here again, the relation *(2)* give us the upper bound
:

*Jß*_{1}exp(ß_{1}) f*(ß_{1})+...+ß_{n}exp(ß_{n}) f*(ß_{n})c ^{p}

with *c* being a constant independant
of *p*.

But, Oh! the two inequality are incompatible for large enough *p*,
hence *i* is transcendantal, so is transcendantal !

We don't really know what just happened, but the result is here!

The original proofs by Hermite and Lindemann are available in *le Petit Archimède*
, but it's even longer and complicated, in fact, it's does not bare any
resemblance of what we did just now...