The reason I said the exact expansion isn't necessary is that the expression for p⁵ in terms of s₁, ..., s₄ for n=4, which must exists, is going to consist of a sum of monomials s_i1^j₁s_i2^j₂...

Now the degree of this must be 5, so for each term we must have

å

ik jk=5

. It is easily seen that since i_k £ 4, each term must involve at least one of s₁ and s₂, and we know these are both zero.

z15 + z25 + z35 + z45 + z55 = 5

å 1 £ i < j < k < l < m £ n

zi zj zk zl zm

. (with a possible minus in front)

This can take any value we want (for n ³ 5). Do you know why?

Suppose we have an arbitrary symmetric polynomial f(x₁,...,x_n). We can divide the terms in the polynomial into sets which are essentially the same monomial under permutations of the variables. I'm going to write [r₁,r₂,...,r_n] for the sum of all terms of the form x₁^r_s(1) x₂^r_s(2) ... x_n^r_s(n)

where the sum is taken over the n! permutations s of (1...n), and without loss of generality r₁ ³ r₂ ³ ... ³ r_n. I shall call a term of this form a monomial (this is slightly different from the standard notation, so beware.)

OK, that probably doesn't make much sense; what I mean is that, for example,

[4,2,1]=x₁⁴ x₂² x₃ + x₁⁴ x₂ x₃² + x₁² x₂⁴ x₃ + x₁² x₂ x₃⁴ + x₁ x₂⁴ x₁² + x₁ x₂² x₁⁴

(Note that by this definition [1,1,1]=6x₁ x₂ x₃; the coefficient actually isn't all that important here.)

Right. Now the first step in the proof is to realise that every symmetric polynomial in x₁ ... x_n can be written as a sum of terms [r₁,r₂,...,r_n] with r₁ ³ r₂ ³ ... etc.

Now, let's put an order on these terms. I will say that [r₁,...,r_n] > [s₁,...,s_n] if r₁ > s₁, or r₁=s₁ and r₂ > s₂, and so on; this is the lexicographic ordering. Crucially it is a total ordering; given r=[r₁ ... r_n] and s=[s₁...s_n], we either have r > s, s > r or r and s are identical. (This is where the proof I found online at PlanetMath.org is wrong: the ordering it defines is not total, and thus entirely useless.) The significance of this total ordering property is that each symmetric polynomial has a unique highest term in this ordering.

Consider an expression of the form e₁ ^t₁ e₂^t₂... e_n^t_n where the e_i's are the elementary symmetric polynomials.

This is evidently itself symmetric. What is the highest term in it (with respect to the above ordering)? It is not too difficult to see that it is in fact [t₁+...+t_n,t₂+...+t_n,...,t_n].

So given an arbitrary symmetric polynomial f, let the highest monomial in it be l[r₁,r₂,...,r_n]. Consider f-le₁^r₁-r₂ e₂^r₂-r₃...e_n^r_n. This is another symmetric polynomial, and crucially the highest monomial in it is lower than [r₁,...,r_n] as we have knocked out this term, and as noted above the highest monomial is unique.

At this point it is tempting to say 'done by induction'. However there is one last point to check: that the monomial ordering is actually a well-ordering, one in which each set has a least member. Otherwise we could keep on inducting downwards infinitely many times without getting anywhere. But this is easy to check as there are only finitely many monomials less that [r₁,...,r_n] for any fixed r₁... r_n. (This is why we needed to adopt the convention that r₁ ³ r₂ etc; otherwise this bit does not hold.) So we really are done. Phew.

e2(x1,...,xn)=x1 x2 + ... + x1 xn + x2 x3 + ... + xn-1 xn=

å 1 £ i1 < i2 £ n

xi1xi2

e3(x1,...,xn)=

å 1 £ i1 < i2 < i3 £ n

xi1 xi2 xi3

...

e_n(x₁,...,x_n)=x₁...x_n

Warning: You might also see these with a multiplicative factor of (-1)ⁿ.