If we make the substitution $\xi = \tau x $ the integration limits $\xi = 0 \rightarrow \tau =0$ and $\xi = x \rightarrow \tau =1$. If we multiply both sides of the equation by $e^{x}$ we get this expression. How does he end up with $kF(0)$? If you start with (9) $$ e^{\beta_1}+...+e^{\beta_r}+k=0 \equiv\\ k=-(e^{\beta_1}+...+e^{\beta_r})\equiv \\ kF(0)= -\sum e^{\beta_i}F(0) $$ How do you prove the limit is zero? $$ \frac{d(e^{-x}F(x))}{dx}=-e^{-x}F(x)+e^{-x}F'(x)=\\ =-e^{-x}F(x)+e^{-x}(F(x)-f(x))=-e^{-x}f(x) $$ In fact it is known that one of those two numbers $e\pi$ or $\pi +e$ is transcendental. Note that if we expand expression (2) we get $$ 1+ ...+e^{\alpha_i+\alpha_j}+e^{\alpha_i+\alpha_j+\alpha_k}+...e^{\overbrace{\alpha_i+...+\alpha_l }^{n}}=0 $$ We showed above that the sum of 2 algebraic numbers is also an algebraic number and so it is possible to create a polynomial $\theta_2(x)$ whose roots are $\alpha_i+\alpha_j$. The same applies to the sum of 3,4,...n algebraic numbers. This is not difficult to see because of the $x^{p-1}$ factor and also because $\theta(\beta_j) = 0$. Yes Mike, for instance if you have the polynomial $f(x)=x^2+1$, the coefficients are rational and the roots are $i$ and $-i$. That's true, but the argument here is just based on the fact that coefficients of a polynomial are (up to a sign) the elementary symmetric functions on its roots. Isn't a $\tau$ missing? I mean, shouldn't the integral be $$\int_0^1\frac{\{c^r\beta_j\tau\theta(\tau\beta_j)\}^p}{\beta_j\tau(p-1)!}e^{(1-\tau)\beta_j}d\tau$$? Luis. Any reference to the latter fact? It's interesting that we know $e$ and $\pi$ are transcendental numbers but no one has proved that $\pi+e$ or $e\pi$ are transcendental. A transcendental number is a real or complex number that is not algebraic - that is, it is not a root of a non-zero polynomial equation with rational coefficients. Check out this video touching on that topic: [](https://www.youtube.com/watch?v=ZxA-xD9mUmM) It is not obvious to me why the resultant is a polynomial in t with rational coefficients. Using expression (11) $$ F(\beta_j)=f(\beta_j)+f^{(1)}(\beta_j)+...+f^{(s+p+1)}(\beta_j) $$ and knowing that $\sum^{r}_{j=1}f^{(t)}(\beta_j)=0$ for $0\leq t<p$ we get $$ \sum^{r}_{j=1} F(\beta_j)=\sum^{r}_{j=1} f(\beta_j)+ \sum^{r}_{j=1} f^{(1)}(\beta_j)+...+ \sum^{r}_{j=1} f^{(s+p+1)}(\beta_j) $$ All the sums where the derivatives have an order $<p$ will be zero. If now we use the result $\sum^{r}_{j=1}f^{(t)}(\beta_j)=pk_t$, we end up with $$ \sum^{r}_{j=1} F(\beta_j)=p\sum^{p+s}_{t=p}k_t $$ We have finally reached a contradiction since the left side of (12) is an integer and the right side is as small as we wish by choosing a very large $p$, which means that $\pi$ cannot be an algebraic number! One way of understanding this limit is thinking about the exponential as $\sum_{n=0}^{\infty}\frac{x^n}{n!}=e^x$. This is a convergent series which means that the weight of the terms with large $n$ will have to tend to zero, otherwise it would diverge. $$ \lim_{n \rightarrow \infty} \frac{x^n}{n!} = 0 $$ This is exactly the same limit we have in the paper if we do $x=c^r\beta_j\theta(\tau\beta_j)$. Is $i$ an algebraic number? Note that $f(x)$ can be written as $$ f(x)=\frac{c^sx^{p-1}(cx^r+...+c_r)^p}{(p-1)!} $$ The highest term in $f(x)$ is of the form $x^{p-1}x^{rp}=x^{p+s}$. $$ f(x)=\frac{c^s(wx^{p+s}+...+w_rx^{p-1})}{(p-1)!} $$ where $w,...,w_r$ are integers that have resulted from expanding the expression $(cx^r+...+c_r)^p$. It's not difficult to see that for any derivative higher than $p$ the coefficients will be divisible by $pc^s$. As shown in a comment above, $f(x)$ can be written as $$ f(x)=\frac{c^s(wx^{p+s}+...+w_rx^{p-1})}{(p-1)!} $$ Since the lowest term in $f(x)$ is of order $p-1$, $f^{(t)}(x)$ for $t<p-1$ will always have terms in $x$ and so $f^{(t)}(0)$ will of course be zero. $f^{(p-1)}(0)=\frac{(p-1)!c^s w_r}{(p-1)!}$ where $w_r$ is the last term in the binomial expansion of $(cx^r+...+c_r)^p$ and so $w_r=c_r^p$. Finally for $t\geq p$, $f^{(t)}(0)$ will only have one term that will not vanish, the term of order $t$. The result is simply a product of $p$, $c^s$ and the binomial expansion term $w_t$. All the terms are integers and so we can group in the form $pK_t$, where $K_t$ in an integer. It's actually not obvious that the product (or sum) of 2 algebraic numbers is also an algebraic number. To prove this let's first define the Resultant $R(f,g)$ of 2 polynomials $f(x)=a_0x^n+...+a_n$ and $g(x)=b_0x^m+...+b_m$ with $a_0b_0 \neq 0$ as $$ R(f,g) = a_0^mb_0^n \prod_{ij}(\alpha_i-\beta_j) $$ where the $\alpha_i$ are the roots of $f(x)$ and $\beta_j$ are the roots of $g(x)$. It's not difficult to notice that if the 2 polynomials f and g have a root in common the resultant will be zero. To prove that the sum of 2 algebraic numbers is also an algebraic number we need to find a polynomial whose roots are $\alpha_i+\beta_j$. To construct such polynomial we will use the resultant of $f(x)$ and $g(t-x)$ $$R(f(x),g(t-x))$$ Now if we do the transformation of variables $x \rightarrow t-x$ the roots of $g(t-x)$ will be $t-\beta_j$ and so $$R(f(x),g(t-x))= a_0^mc_0^n \prod_{ij}(\alpha_i-(t-\beta_j))=\\ =a_0^mc_0^n \prod_{ij}(\alpha_i+\beta_j-t) $$ Now, it's not difficult to see that the resultant itself will be a polynomial in $t$ of degree $mn$ with roots $\alpha_i+\beta_j$. To prove that the product of 2 algebraic numbers is also an algebraic number we use a different resultant: $$ R(f(x),x^mg(t/x)) $$ $$ x^mg(1/x)=b_0+...+b_mx^m $$ The roots of $x^mg(1/x)$ are $1/\beta_j$ and so the roots of $x^mg(t/x)$ are $t/\beta_j$. If we now plug that into the resultant we get a polynomial with degree $mn$ in t whose roots obey the equation: $$ \alpha_i-t/\beta_j=0\\ t=\alpha_i\beta_j $$ Ivan Niven was a Canadian-American mathematician, specializing in number theory. He is famous for completing the solution of most of Waring's problem in 1944. He also has an Erdős number of 1 because he has coauthored a paper with Paul Erdös.