Proof: Since $\underbrace{11 \cdots 1}_{n \uparrow}=\frac{10^{n}-1}{9}$, and
12345678987654321
is the sum of the following numbers:
11111111111111111
1111111111111110
111111111111100
11111111111000
1111111110000
111111100000
11111000000
1110000000
100000000
Therefore, 12345678987654321
$$\begin{aligned} = & \frac{10^{17}-1}{9}+10 \times \frac{10^{15}-1}{9}+10^{2} \times \frac{10^{13}-1}{9}+ \\ & \cdots+10^{8} \times \frac{10^{1}-1}{9} \\ = & \frac{1}{9}\left[\left(10^{17}-1\right)+\left(10^{16}-10\right)+\left(10^{5}-10^{2}\right)+\right. \\ & \left.\cdots+\left(10^{9}-10^{8}\right)\right] \\ = & \frac{1}{9}\left[\left(10^{17}+10^{16}+10^{15}+\cdots+10^{9}\right)-\right. \\ & \left.\left(1+10+10^{2}+\cdots+10^{8}\right)\right] \\ = & \frac{1}{9}\left[10^{9}\left(10^{8}+10^{7}+10^{6}+\cdots+10+1\right)-\right. \\ & \left.\left(1+10+10^{2}+\cdots+10^{8}\right)\right] \\ = & \frac{1}{9}\left(10^{9}-1\right)\left(1+10+10^{2}+\cdots+10^{8}\right) \\ = & \frac{1}{9}\left(10^{9}-1\right) \frac{10^{9}-1}{9}=\left(\frac{10^{9}-1}{9}\right)^{2} \\ = & \underbrace{11 \cdots 1^{2}}_{91} . \end{aligned}$$

Therefore, 12345678987654321 is a perfect square.

For this problem, my student Chen Chang provided a simpler proof:
Let $A=\underbrace{11 \cdots 1}_{9 \uparrow}$. Then
12345678987654321
$$\begin{array}{l} =A+10 A+10^{2} A+\cdots+10^{8} A \\ =\underbrace{11 \cdots 1}_{9 \uparrow} \times A=A^{2} . \end{array}$$

Actually, 12345678987654321 is the sum of the following numbers:
111111111
1111111110
11111111100
111111111000
1111111110000
11111111100000
111111111000000
1111111110000000
11111111100000000
Therefore, 12345678987654321 is a perfect square.
Note: A related interesting pyramid is as follows:
$$\begin{aligned} 1^{2} & =1 \\ 11^{2} & =121 \\ 111^{2} & =12321 \\ 1111^{2} & =1234321 \\ 11111^{2} & =123454321 \\ 111111^{2} & =12345654321 \\ 1111111^{2} & =1234567654321 \\ 11111111^{2} & =123456787654321 \\ 111111111^{2} & =12345678987654321 \end{aligned}$$