7.8 Let $m, n, p, q$ be non-negative integers, and for all $x>0$,
$$
\frac{(x+1)^{m}}{x^{n}}-1=\frac{(x+1)^{p}}{x^{q}}
$$

always holds, find the value of $\left(m^{2}+2 n+p\right)^{2 q}$.
[Solution] Since
$$
\frac{(x+1)^{m}}{x^{n}}-1=\frac{(x+1)^{p}}{x^{q}}
$$

holds for all $x>0$, we can take $x=1$, obtaining
$$
2^{m}-1=2^{p} \text {. }
$$

Since $m$ and $p$ are non-negative integers, $2^{m}-1$ is a positive odd number, so it can only be that $2^{p}=1, p=0$.
Thus,
$$
2^{m}-1=1, m=1 .
$$

Now, taking $x=2$ in equation (1) and noting that $m=1, p=0$, we have
$$
\begin{array}{c}
\frac{3}{2^{n}}-1=\frac{1}{2^{q}} . \\
3 \cdot 2^{q}-2^{n+q}=2^{n} . \\
3=2^{n-q}+2^{n} .
\end{array}
$$

If $n>q$, then the left side of the above equation is odd, while the right side is even, which is impossible; if $n<q$, then the left side is an integer, while the right side is a fraction, which is also impossible; thus, we must have $n=q$. Therefore, $3=2^{0}+2^{n}$, which implies
$$
n=1, q=1 \text {. }
$$

Hence, $\left(m^{2}+2 n+p\right)^{2 q}=(1+2+0)^{2}=9$.