EDIT: This is a case of being too wrapped up in a formulation
($e\_j,p\_i,$ and the like) to try something simple. It did not
occur to me to pull exp to the outside in the weeks I have 
stared at this. Thanks to Aaron Meyerowitz and Brendan McKay for
a humbling revelation. I leave it to others to decide if this is
an object lesson in mathematical myopia or a question to be removed.
END EDIT


If the following turns out to be a homework question, great!
Just tell me what course, textbook (or online accessible .pdf)
and page number has the problem, and I'll go check it out.


Let $m$ and $k$ be integers with $k \gt e^m \geq 1$. I am 
considering two expressions in variables $x\_i$, and looking
at the partial sums with monomials of degree at most $m$,
which I note by $\cong\_m$. (So if any term like, say, $x\_hx\_i^{m-1}x\_j$ occurs
in my sum, I toss that term and consider what is left with $\cong\_m$.)
 The question in brief is:
$$\prod\_{0 \leq i \leq k}(1 - x\_i) \cong\_m
\prod\_{1 \leq j}(\sum\_{0 \leq l} [ - \sum\_{0 \leq i \leq k}(x\_i)^j/j ]^l/l! ) ?$$


Here is some background and motivation. I rewrite the above question in terms
of elementary symmetric polynomials in the $x\_i, 0\leq i \leq k$.  

In order to have enough terms to make sense, assume $k$ is as large as needed.
(Likely I just need $k > e^{e^m}$, but it would be nice to know if, say, just 
$k \gt m^2$ will do.) Let us write 
$q\_j$ for $ \sum\_{0 \leq i \leq k} (x\_i)^j/j $. This is $p\_j/j$, or the
$jth$ power symmetric polynomial divided by $j$. Let us also write $XP\_m(t)$ for some
polynomial $t$ to be the formal expression (which we will chop off after the
terms get too big) $\sum\_{0\leq j \leq m} (-t)^j/j!$, after the series expansion
for $e^{-t}$. For the left hand side, we rewrite it in terms of the
elementary symmetric polynomials $e\_j$, which is the sum over all $j$-sets
(not $j$-tuples) $\{i\_1, \ldots , i\_j\}$ of $\{0,\ldots,k\}$ of the monomials
$x\_{i\_1}\ldots x\_{i\_j}$. I also declare $e\_0=1$.
 Now the left hand side is written as a sum, and
as I am concerned only with that part containing monomials of total degree at most $m$,
I can limit the summation indices and 
the question now becomes:


$$\sum\_{ 0 \leq i \leq m} (-1)^ie\_i \cong\_m \prod\_{1\leq j \leq m} XP\_m(q\_j) ?$$


If I had facility with any computer algebra system, I would have tried it there first
before asking here. I have verified the equality
for small $m$, and would like a reference to a proof, or disproof.
Also, Newton's identities $je\_j = \sum\_{1\leq i \leq j}(-1)^{i-1}p\_ie\_{j-i}$ give
me some hope that the above is true, but I have not found a derivation.


My motivation is in exploring the limits of an argument I learned in a paper
of Harlan Stevens. He uses the left hand side with prime reciprocal values for the
$x\_i$, and looks for the smallest odd integer m such that the sum is positive,
but all partial sums for smaller odd positive integers are negative, so in
particular $q\_1=p\_1\gt 1$ for interesting examples.  

If it is true, I hope to show that the smallest odd positive $m$ that gives
a positive partial sum implies $p\_1 \gt Cm$ for some real constant $C$, which would
then tell me that this argument is limited to showing upper bounds no tighter than $Ak^{B\log\log k}$,
and no smaller.


Gerhard "Also, I Find It Pretty" Paseman, 2014.01.19



Truncated sums of symmetric polynomials; reference request for an algebraic derivation
Try this:
 $$\prod\_{1 \leq j}^{\infty}(\sum\_{0 \leq l}^{\infty} [ - \sum\_{i=0}^k\frac{x\_i^j}{j} ]^l/l! ])=$$
$$\prod\_{1 \leq j}^{\infty}\mathop{exp}\left( - \sum\_{i=0}^k\frac{x\_i^j}{j}\right)=$$
$$\mathop{exp}\left(\sum\_{j=1}^{\infty}\left( - \sum\_{i=0}^k\frac{x\_i^j}{j}\right)\right)=$$
$$\mathop{exp}\left(\sum\_{i=0}^k\left( -\sum\_{j=1}^{\infty} \frac{x\_i^j}{j}\right)\right)=$$
$$\mathop{exp}\left(\sum\_{i=0}^k\ln({1-x\_i)}\right)=$$
$$=\prod\_{i=0}^k(1-x\_i
)$$


So I think $m=k$ would be just enough.


