7. [5] Simplify the product
$$
\prod_{m=1}^{100} \prod_{n=1}^{100} \frac{x^{n+m}+x^{n+m+2}+x^{2 n+1}+x^{2 m+1}}{x^{2 n}+2 x^{n+m}+x^{2 m}}
$$

Express your answer in terms of $x$.
Answer:
$$
x^{9900}\left(\frac{1+x^{100}}{2}\right)^{2}\left(\text { OR } \frac{1}{4} x^{9900}+\frac{1}{2} x^{10000}+\frac{1}{4} x^{10100}\right)
$$

Solution: We notice that the numerator and denominator of each term factors, so the product is equal to
$$
\prod_{m=1}^{100} \prod_{n=1}^{100} \frac{\left(x^{m}+x^{n+1}\right)\left(x^{m+1}+x^{n}\right)}{\left(x^{m}+x^{n}\right)^{2}} .
$$

Each term of the numerator cancels with a term of the denominator except for those of the form $\left(x^{m}+x^{101}\right)$ and $\left(x^{101}+x^{n}\right)$ for $m, n=1, \ldots, 100$, and the terms in the denominator which remain are of the form $\left(x^{1}+x^{n}\right)$ and $\left(x^{1}+x^{m}\right)$ for $m, n=1, \ldots, 100$. Thus the product simplifies to
$$
\left(\prod_{m=1}^{100} \frac{x^{m}+x^{101}}{x^{1}+x^{m}}\right)^{2}
$$

Reversing the order of the factors of the numerator, we find this is equal to
$$
\begin{aligned}
\left(\prod_{m=1}^{100} \frac{x^{101-m}+x^{101}}{x^{1}+x^{m}}\right)^{2} & =\left(\prod_{m=1}^{100} x^{100-m} \frac{x^{1}+x^{m+1}}{x^{1}+x^{m}}\right)^{2} \\
& =\left(\frac{x^{1}+x^{1} 01}{x^{1}+x^{1}} \prod_{m=1}^{100} x^{100-m}\right)^{2} \\
& =\left(x^{\frac{99 \cdot 100}{2}}\right)^{2}\left(\frac{1+x^{100}}{2}\right)^{2}
\end{aligned}
$$
as desired.