You can remove almost all of the words in your proof! By including so many details, you make the proof hard to read. This is all you need:
Let $a,b \in \mathbb{R}$ s.t. $a<b$. By the density of the rational numbers in the reals, $\exists\, q\neq0 \in \mathbb{Q}$ s.t. $\frac{a}{\pi} < q < \frac{b}{\pi}$ and therefore $a < \pi q < b$. Since $\pi q$ is transcendental, we are done.
If your reader doesn't believe that $\frac{a}{\pi} < q < \frac{b}{\pi}$ implies $a < \pi q < b$, or that $\pi q$ is transcendental when $q\neq0$, then it's their responsibility to educate themselves.
You had better take a look at Ivan Niven's book Irrational Numbers where this technique of Hermite is described in a simple manner and used to prove many results apart from transcendence of $e$.
He presents the following key result (Lemma 2.3 in his book on page 16):
Theorem: If $h(x) = x^mg(x) / m! $ where $g(x) $ is a polynomial with integer coefficients and $m$ is a positive integer, then $h^{(j)} (0)$, the $j$-th derivative of $h(x) $ at $x=0$, is an integer for $j=0,1,2,\dots$. Moreover, with the possible exception of the case $j=m$, the integer $h^{(j)} (0)$ is divisible by $m+1$; no exception need be made in the case $j=m$ if $g(x) $ has the factor $x$: ie, if $g(0)=0$.
Your $f(x) $ is $h(x) $ with $m=p-1$ with $p$ a prime and $$g(x) = (1-x)^p(2-x)^p\dots(m-x)^p$$ The above theorem then implies that all derivatives of $f$ at $0$ are integers and all of them are divisible by prime $p$ except possibly the $(p-1)$-th derivative. Further the value $p$ is chosen to ensure that $f^{(p-1)}(0)=(n!)^{p}$ is not divisible by $p$ (this is done by making $p>n$).
Further $$F(x) =\sum_{j=0}^{(n+1)p-1}f^{(j)}(x)$$ and $G(x) =e^{-x} F(x) $ so that $G'(x) =-e^{-x} f(x) $. Ivan Niven makes use of integrals from this point onwards but the same can be achieved by using mean value theorem (this is the approach in your question and by I N Herstein in his Topics in Algebra).
We have $$e^{-k} F(k) - F(0)=-ke^{-x_k}f(x_k)$$ or $$F(k) - e^kF(0)=-ke^{k-x_k}f(x_k)$$ Multiplying the above equation by $c_k$ and adding such equations for $k=1,2,\dots,n$ we get $$\sum_{k=1}^{n}c_kF(k)-F(0)\sum_{k=1}^{n}c_ke^k=-\sum_{k=1}^{n}kc_ke^{k-x_k}f(x_k)$$ By assumption we have $\sum_{k=1}^{n}c_ke^k=-c_0$ and therefore the last equation above can be written as $$\sum_{k=0}^{n}c_kF(k)=-\sum_{k=1}^{n}kc_ke^{k-x_k}f(x_k)\tag{1}$$ Now we do careful analysis of the expression on left side of the above equation and conclude that it is an integer not divisible by prime $p$ and is therefore non-zero (this also requires $p>|c_0|$).
The analysis of the term $c_0F(0)$ is a direct application of Niven's theorem and it is an integer not divisible by $p$ because $f^{(j)} (0)$ is divisible by $p$ for all $j\neq p-1$ and $f^{(p-1)}(0)$ is not divisible by $p$.
For analysis of $c_kF(k), k\neq 0$ note that if $p_k(x) =f(k-x) $ then $$p_k^{(j)} (0)=(-1)^{j}f^{(j)}(k)$$ and further $p_k(x) = f(k-x)$ can be written as $$\frac{x^{p-1}}{(p-1)!}xP_k(x)$$ for some polynomial $P_k(x) $ with integer coefficients. Therefore by Niven's theorem all the derivatives $p_k^{(j)} (0)$ and hence $f^{(j)} (k) $ are divisible by $p$. And therefore $c_kF(k) $ is an integer divisible by $p$.
Next step is the estimation of right side of $(1)$ and here one notices that $$|f(x_k)|\leq \frac{n^{(n+1)p-1}}{(p-1)!}$$ and therefore the right side of $(1)$ can be made arbitrarily small (in particular less than $1$) in absolute value by choosing large prime $p$ and we then get a contradiction.
Best Answer
One could use the growth at infinity of the function $f:x\mapsto\mathrm e^x$.
Assume that that $f$ is algebraic and choose a real number $x\geqslant0$. Then $|f(x)|\geqslant1$ and $$ |c_n(x)|\,|f(x)|^n\leqslant b(x)|f(x)|^{n-1},\qquad b(x)=\sum\limits_{k=0}^{n-1}|c_k(x)|. $$ Hence, for every real number $x\geqslant0$ such that $c_n(x)\ne0$, $|f(x)|\leqslant b(x)/|c_n(x)|$. But indeed, $c_n(x)\ne0$ for every real number $x$ large enough and $b(x)/|c_n(x)|$ can grow at most polynomially when the real number $x$ goes to $+\infty$ while $|f(x)|=\mathrm e^x$ grows... well, exponentially. This is a contradiction.