You can actually prove e is transcendental in a way fairly similar to the submitted proof that pi is irrational, just with a lot more fiddling. There's a proof at the end of chapter 2 of Niven's "Irrational Numbers", which is available on the Internet Archive. I'll summarize below.
Assume e is algebraic of degree m, with A_m e^m + A_(m-1) e^(m-1) + ... + A_0 = 0 where the A_i's are integers, A_0 != 0.
1. Define a polynomial f(x) = x^(p-1) (x-1)^p (x-2)^p ... (x-m)^p / (p-1)! where p is an odd prime to be specified later.
2. Then define F(x) as f(x) + f'(x) + f^(2)(x) + ... + f^(mp+p-1)(x).
3. Show that for 0 < x < m we have |f(x)| < m^(mp+p-1)/(p-1)!.
4. Show that the derivative of e^(-x) F(x) = -e^(-x) f(x).
5. Show that A_j integral( e^(-x) f(x), 0, j) = A_j F(0) - A_j e^(-j) F(j).
6. Multiple that by e^j and then sum over j = 0, 1, ..., m giving sum( A_j e^j integral(e^(-x)f(x), 0, j), j = 0 -> m) = - sum( sum( A_j f^(i)(j), i= 0 -> mp+p-1), j= 0 -> m).
7. Show that f^(i)(j) is an integer for all i, j in that sum, and that each is divisible by p except in the case j = 0, i = p-1.
8. Show that f^(p-1)(0) is not divisible by p if we choose p > m.
9. Show that if we chose p > |A_0| the double sum on the right in #6 consists of multiples of p except for for the term -A_0 f^(p-1)(0), and so must be a non-zero integer.
10. Use the inequality from #3 in the left side in #6 to conclude that |left side| < 1 if p is sufficiently large. That contradicts #9.
Assume e is algebraic of degree m, with A_m e^m + A_(m-1) e^(m-1) + ... + A_0 = 0 where the A_i's are integers, A_0 != 0.
1. Define a polynomial f(x) = x^(p-1) (x-1)^p (x-2)^p ... (x-m)^p / (p-1)! where p is an odd prime to be specified later.
2. Then define F(x) as f(x) + f'(x) + f^(2)(x) + ... + f^(mp+p-1)(x).
3. Show that for 0 < x < m we have |f(x)| < m^(mp+p-1)/(p-1)!.
4. Show that the derivative of e^(-x) F(x) = -e^(-x) f(x).
5. Show that A_j integral( e^(-x) f(x), 0, j) = A_j F(0) - A_j e^(-j) F(j).
6. Multiple that by e^j and then sum over j = 0, 1, ..., m giving sum( A_j e^j integral(e^(-x)f(x), 0, j), j = 0 -> m) = - sum( sum( A_j f^(i)(j), i= 0 -> mp+p-1), j= 0 -> m).
7. Show that f^(i)(j) is an integer for all i, j in that sum, and that each is divisible by p except in the case j = 0, i = p-1.
8. Show that f^(p-1)(0) is not divisible by p if we choose p > m.
9. Show that if we chose p > |A_0| the double sum on the right in #6 consists of multiples of p except for for the term -A_0 f^(p-1)(0), and so must be a non-zero integer.
10. Use the inequality from #3 in the left side in #6 to conclude that |left side| < 1 if p is sufficiently large. That contradicts #9.