My gut feeling is that this shouldn't typically be the case, but I'm tired and struggling to convey why without descending into downright criminal levels of vagueness. I'd be more hopeful about an algebraic number of degree >1 having a 'nice' reason for specifically being irrational.
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.