(Relatively) easy Laplace equation on a square

Solve Laplace's equation $\Delta u=0$ on the unit square subject to the boundary conditions
$u(x,0)=1$ and $u(1,y)=u(x,1)=u(0,y)=0$.

The answer here for the basic Dirichlet problem on the unit square is $$u(x,y)= \sum_{n=1}^\infty (c_ne^{-n\pi y}+d_ne^{n\pi y})sin(n\pi x)$$. Where the coefficients $c_n+d_n=2\int_{0}^{1}(1)sin(n\pi x)dx$ and $c_ne^{-n\pi}+d_ne^{n\pi}=\int_{0}^{1}0dx=0$. The first fact gives us $$2(\frac{-cos(n\pi x)}{n\pi}|_0^1)=\frac{2(1-(-1)^n)}{n\pi}$$. The second fact gives us that $$d_n=\frac{-c_ne^{-n\pi}}{e^{n\pi}}=-c_ne^{-2n\pi}$$. Substituting into the first equation gives: $c_n-c_ne^{-2n\pi}=\frac{2(1-(-1)^n)}{n\pi}$ Thus $$c_n(1-e^{-2n\pi})=\frac{2(1-(-1)^n)}{n\pi};\>c_n=\frac{2(1-(-1)^n)}{n\pi(1-e^{-2n\pi})}$$. It follows that $$d_n=\frac{2(1-(-1)^n)}{n\pi}-c_n=\frac{2(1-(-1)^n)}{n\pi}-\frac{2(1-(-1)^n)}{n\pi(1-e^{-2n\pi})}$$. Therefore the final answer to the problem is $u(x,y)=$ $$\sum_{n=1}^{\infty} \{(\frac{2(1-(-1)^n)}{n\pi(1-e^{-2n\pi})})e^{-n\pi y}+(\frac{2(1-(-1)^n)}{n\pi}-\frac{2(1-(-1)^n)}{n\pi(1-e^{-2n\pi})})e^{n\pi y}\}sin(n\pi x) $$ $\blacksquare$

I'm having a little trouble following some of your stuff, but I also did things in a different order. I used $u(x,1)=0$ to find that $c_n=-d_ne^{2n\pi}$ (which I followed up to that point with yours. Then I only had one coefficient to worry about
$$u(x,y)=\sum_{n=1}^\infty d_n(e^{n\pi y}-e^{-n\pi y}e^{2n\pi})\sin(n\pi x)$$
Which after applying $u(x,0)=1$ gave me
$$\begin{array}u(x,y)&=\sum_{n=1}^\infty \frac{2}{n\pi}\frac{1-(-1)^n}{1-e^{2n\pi}}(e^{n\pi y}-e^{-n\pi y}e^{2n\pi})\sin(n\pi x)\\
&=\sum_{n=1}^\infty 2 \frac{1-(-1)^n}{n\pi}\frac{1-e^{2n\pi(1-y)}}{1-e^{2n\pi}}e^{n\pi y}\sin(n\pi x)
\end{array}$$

I think these are equal so I think I agree with your final answer. Though I am confused about why you have an end of proof symbol.