Finding the degree of minimal polynomials Let a number $x = \sqrt[a_1]{p_1} + \sqrt[a_2]{p_2} + \ .. \ + \sqrt[a_n]{p_n}$ be a number such that all $a_n$ are integers and all $p_n$ are rational. I've been noticing that for every number x, the degree of its minimal polynomial is seemingly always equal to $\prod_{1}^n \ a_n$.
Is that valid for all values of $a_n$? If so, is there a proof?
 A: No. Some conditions are needed on the $a_i$ and $p_i$. For instance, take n=2, $a_1 = a_2 = 2$, $p_1 = p_2 = 2$. Then $x = 2 \sqrt{2}$, which has minimal polynomial $x^2 - 8$. As an even simpler example, n=1, $a_1 = 2$, $p_1 = 4$, then $x$ is rational.
For a less trivial example, take $a_1= 4$, $a_2 = 6$, $p_1=p_2=2$. Check that this has a polynomial of degree 12. In fact, this isn't really true at all. 
One can, however, prove that the degree of the minimal polynomial is at most $\prod a_n$, which is an easy exercise in field theory. Any graduate algebra textbook covering Galois theory will be more than sufficient to prove this; just remember the degree of the minimal polynomial is the same as the dimension of the extension field viewed as a vector space over the base field.
EDIT:
After much miscommunication on my part, we've reached the following results:
Suppose $a_1,\ldots,a_n$ are pairwise relatively prime positive integers, $p_1, \ldots, p_n$ integers such that $\sqrt[a_i]{p_i}$ is of degree $a_i$ for each i. Then $\sqrt[a_1]{p_1} + \cdots + \sqrt[a_n]{p_n}$ is of degree $\displaystyle \prod_{i=1}^n a_i$.
The condition that each $\sqrt[a_i]{p_i}$ is met (by Eisenstein Criterion) should there be a prime $q_i$ such that $q_i | p_i$ and $q_i^2 \not{|} p_i$ for each i.
A: Besicovitch has proved the following related interesting result:

Consider an integer $n\gt 1$ and distinct prime numbers $p_1,p_2,\ldots ,p_k.$ Then the field $F=\mathbb Q (\sqrt[n]{p_1},\ldots ,\sqrt[n]{p_k})$ 
  has dimension $n^k$ over $\mathbb Q$ .
  More precisely, a $\mathbb Q$-basis of that field $F$ is given by the radicals
  $$\sqrt[n]{p_1^{m_1}\ldots p_i^{m_i} \ldots p_k^{m_k} } \quad (\; 0\leq m_i \lt n \quad , \quad 1\leq i\leq k )     $$ 

(The case $n=2$ is a classical chestnut in Galois theory.)
This does not answer the OP's question but at least assures us that, for example,
$$\sqrt[3]{900}+\sqrt[3]{36}+ \sqrt[3]{15}+\sqrt[3]{150} \notin \mathbb Q $$
 which is not so simple to check directly.        
I have the pessimistic feeling that there is no very satisfactory general answer to the question "when does the sum 
$ \sqrt[n_1]{a_1}+ \sqrt[n_2]{a_2}+...+\sqrt[n_k]{a_k}$ have degree $n_1 n_2 ...n_k$", but I'd love to be shown wrong.
Bibliography: Besicovich's original article is:   Abram S. Besicovitch, "On the linear independence
of fractional powers of integers", Journal of the
London Mathematical Society 15 (1940), 3-6.
Here is a more recent and accessible proof : Ian Richards, "An application of Galois theory
to elementary arithmetic", Advances in Mathematics 13 (1974), 268-273.
13 (1974), 268-273.
A: The canonical references for this are:
MR0818878 (87b:68058) 
Zippel, Richard(1-MIT-C)
Simplification of expressions involving radicals. 
J. Symbolic Comput. 1 (1985), no. 2, 189–210. 
MR1148819 (92k:12008) 
Landau, Susan(1-MA-C)
Simplification of nested radicals. 
SIAM J. Comput. 21 (1992), no. 1, 85–110. 
and more recently
MR1776235 (2001g:12004) 
Blömer, J.(D-PDRB)
Denesting by bounded degree radicals. (English summary) 
Fifth European Symposium on Algorithms (Graz, 1997). 
Algorithmica 28 (2000), no. 1, 2–15. 
