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I'm trying to get through the proof of transcendence of $e$ (the base of the natural logarithm) already for a couple of days, but now I got seriously stuck.
Proof is in most sources roughly the same. I followed this version.

We prove the transcendence of $e$ by contradiction - suppose that $e$ is an algebraic number. That means there is a nonzero polynomial $P \in \mathbb{Z}[x]$ such that $P(e) = 0$.
Let $P(x) = \sum_{i=0}^{n}w_{i}x^{i}.$

Lemma
Let $f\in \mathbb{R}[x]$ be a polynomial, $t\in \mathbb{R}^{+}$. Then the following equality holds: $ \int_{0}^{t} e^{-x}f(x)\text{d}x = \sum_{i=0}^{\infty}f^{(i)}(0) - e^{-t}\sum_{i=0}^{\infty}f^{(i)}(t)$

Lemma can be proved by integration per partes. For clarity let's define $F(x) = \sum_{i=0}^{\infty}f^{(i)}(x)$.
Now the lemma looks like this:

$ \int_{0}^{t} e^{-x}f(x)\text{d}x = F(0) - e^{-t}F(t)$

$ \int_{0}^{t} e^{t-x}f(x)\text{d}x = e^{t}F(0) - F(t)$

Because $e$ is (by contradiction) an algebraic number, we have $\sum_{i=0}^{n}w_{i}e^{i} = 0$ for some $w_{i}\in \mathbb{Z}$. Now a rather complicated step. Write the last integral identity for $t = 0$, $1$, ..., $n$, multiply each of them by $w_{t}$ and add the results.
You should obtain: $ \sum_{i=0}^{n}w_{i}\int_{0}^{i} e^{i-x}f(x)\text{d}x = F(0)\sum_{i=0}^{n}w_{i}e^{i} - \sum_{i=0}^{n}w_{i}F(i) $ See why we did this? Using the property of an algebraic number mentioned above we get: $ \sum_{i=0}^{n}w_{i}F(i) = - \sum_{i=0}^{n}w_{i}\int_{0}^{i} e^{i-x}f(x)\text{d}x $ Now the idea of the proof is to choose the polynomial $f$ so wisely that the left side is a non-zero integer and the right side is small (say, less than $\frac{1}{10}$ in absolute value), which gives the contradiction. The wisely chosen polynomial is $f(x) = \frac{1}{(p-1)!}x^{p-1}\prod_{i=1}^{n}(x-i)^p$, where $p\in \mathbb{P}$ is some prime that will be specified later.

And here comes my digging. I am brave, I don't need such a giant polynomial to conclude the contradiction. My chosen polynomial will be $f(x)=x$. Now my equality looks like this: $\sum_{i=0}^{n}w_{i}(i+1) = - \sum_{i=0}^{n}w_{i}\int_{0}^{i} e^{i-x}x\text{d}x$. Uhh, no contradiction. :( Maybe wrong polynomial. Okay, let's take a general polynomial f(x). Now let's compute that integral and find out whether there's a contradiction or not: $ \sum_{i=0}^{n}w_{i}F(i) = - \sum_{i=0}^{n}w_{i}\int_{0}^{i} e^{i-x}f(x)\text{d}x $ $ \sum_{i=0}^{n}w_{i}F(i) = - \sum_{i=0}^{n}w_{i}(e^{i}F(0) - F(i)) $ $ 0 = 0 $ Damn! Neither for $f(x)=x$ nor for $f(x) = x^{2}$ there's no contradiction. Okay, that's understandable - the wisely chosen polynomial wouldn't be so giant if it didn't have to be. But now I showed that when I try to substitute any possible polynomial and compute the integral, it doesn't lead to contradiction! Can you help me? What am I missing here?

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    @fedja "you just effectively cancelled your initial step" Ah, now I see that. It all started with those easy polynomials. I guess that the trick is to do whatever but __not to evaluate that integral__, which always leads back. That might be the reason why the proof continues with establishing the upper bound for the integral instead of evaluating it.2012-11-05

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