The general formula is
$$\left( {\begin{array}{*{20}c} 2N \\ N \\ \end{array}} \right) = \dfrac{{(2N)!}}{{N! (2N-N)!}}\; \hspace{15pt}\leftarrow \hspace{15pt} \tiny \textcolor[rgb]{1,,0}{\text {(Corrected error in binomial formula)}}\\$$
This works for all square grids from N=1 to any finite number.
A single square is
$$\left( {\begin{array}{*{20}c} 2(1) \\ 1 \\ \end{array}} \right) = \; 2\\$$
(one unit up and one unit right, or one unit right and one unit up).
This can be extended to a non-square grid of (R * C) where R and C are rows and columns.
For this the general formula is
$$\left( {\begin{array}{*{20}c} R+C \\ C \\ \end{array}} \right)\;=\;
\dfrac{{(R + C)!}}{{R!\; C!}} \;=\; \text {unique\;paths}
\hspace{15pt}\leftarrow \hspace{15pt} \tiny \textcolor[rgb]{1,,0}{\text {(Corrected \& clarified )}}\\$$
I lack the skills to demonstrate a formal maths proof. Here’s the basic logic:
In a square grid of size N when the number of units between diagonal corners is set to equal 2N, half the route will compose the X direction and half will compose the Y direction. The position of the square (X or Y) determines the position of subsequent squares (X or Y). The value of interest is the unique routes consisting of equal X and Y directions, this is found by “choosing the N positions from the 2N that are available.
(The solutions are close to the number of ways to catch CDD).
$$\tiny \textcolor[rgb]{1,,0}{\text {(Edited: Corrected errors \& clarified )}}\\$$
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