Prove the transcendence of the number $e$ How to prove that the number $e=2.718281...$ is a transcendental number?   The truth is I have no idea how to do it. 

If I can recommend a book or reference on this topic thank you.

There are many tests on the transcendence of $ e $?   

I'd read several shows on the transcendence of $ e $
Try Michael Spivak's Calculus. I find it amusing that he would prove the transcendence of $ e $ in a calculus textbook.
I realize that amWhy already gave one answer: the Lindemann-Weierstrass theorem.  I don't know how to prove this... to be frank it is over my head.  But in general it states that in the equation:
$$e^a = b$$
If $a\ne 0$ is algebraic then $b$ is transcendental. The contrapositive is also true: if $b$ is algebraic then $a$ must be transcendental. Obviously $a=0,b=1$ has to be an exception.

Take that on faith for a moment. Like I said, I cannot prove the L-W theorem but I welcome you to research it.

If you ask whether or not $\ln(2)$ is algebraic or transcendental, take note that $e^{\ln(2)}=2$ is algebraic, and you can conclude that $\ln(2)$ is transcendental.

Similarly, you asked if $e$ itself was algebraic or transcendental. That is easy. $e^1 = e$ by definition. The power is algebraic therefore $e$ is transcendental. This is not circular reasoning. It rests squarely on the truth of the Lindemann-Weierstrass theorem: for any non-zero algebraic power, $e^a$, whatever that may be, is transcendental.

If you ever see a proof for the transcendence of $e$ more complicated than the one I just gave, I assure you it is just a proof for a special case of the L-W theorem. The general case proof would suit you better.

Likewise, take Euler's Identity: $e^{i\pi} = -1$.  The result is algebraic, therefore the power $i\pi$ is transcendental. But we already know $i$ to be algebraic, so it is the $\pi$ that is transcendental. This is how you prove the transcendence of $\pi$.

I would also like to mention a theorem that hasn't been mentioned yet: the Gelfond-Schneider theorem.  It states that:
$$a^b$$
is always transcendental if:


$a$ is algebraic and $a\ne 0,1$, and
$b$ is algebraic and irrational


Numbers such as $2^{\sqrt{2}}$ are proven transcendental in this way.

We can also prove that $\sqrt{2}^{\sqrt{2}}$ is transcendental as a consequence of the G-S theorem, because it is simply the square root of $2^{\sqrt{2}}$. The square root of a transcendental number is still transcendental.

I, personally, do not know of any other helpful theorems. But then I am new to transcendental number theory.  Good luck.