Combinatorial proof that binomial coefficients are given by alternating sums of squares?

More a visual proof than combinatorial, perhaps:

$$ 5^2 - 4^2 + 3^2 -2^2 +1^2 = {6 \choose 2}$$

On the left, you have the alternating sum as an inclusion-exclusion of squares: the total sum is the number of coloured cells.

On the right, you have those L shaped shapes rearranged in the top left of a 6x6 grid. If you think of each cell as a coordinate $(x_1,x_2)$ that gives two elements chosen from the set $\{1, 2 \cdots 6\}$, it's seen that the elements are choosen with $ x_2 > x_1$, what corresponds to a combination (no repetition, and no order).

Added: The others are well known, but, just for the sake of completeness...

$$\displaystyle \sum_{k=1}^n (-1)^{n-k} k^2 = {n+1 \choose 2} = \sum_{k=1}^n \; k = \frac{(n+1) \; n}{2}$$

Get rid of the negative signs by putting together the two last terms of the sum, then the two last before these, and so on. The result is the area of a subset of the $n\times n$ square made of $1\times1$ squares composing L-shaped strips: the largest strip is the union of the $n$ column and the $n$ horizontal line, consecutive strips are separated by similar strips, each strip starts from the vertical axis of coordinates then goes East then goes South and ends at the horizontal axis of coordinates, and I hope that with these indications everybody can picture the result.

Now, color in black the strips composing the subset one wants to measure and color in white the rest of the rectangle:

$\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare$
$\Box\Box\Box\Box\blacksquare$
$\blacksquare\blacksquare\blacksquare\Box\blacksquare$
$\Box\Box\blacksquare\Box\blacksquare$
$\blacksquare\Box\blacksquare\Box\blacksquare$

Next, add a white line of width $n$ to the top of the picture:

$\Box\Box\Box\Box\Box$
$\blacksquare\blacksquare\blacksquare\blacksquare\blacksquare$
$\Box\Box\Box\Box\blacksquare$
$\blacksquare\blacksquare\blacksquare\Box\blacksquare$
$\Box\Box\blacksquare\Box\blacksquare$
$\blacksquare\Box\blacksquare\Box\blacksquare$

The resulting rectangle has size $n\times(n+1)$ and I pretend that it is evenly colored. This will prove the result since its area is $n(n+1)$.

To show this, one can pair together sub-rectangles of opposite colors. The first pair is made of the whole line $n+1$ (white) and the whole line $n$ (black). Erase them both. The resulting rectangle has size $n\times(n-1)$:

$\Box\Box\Box\Box\blacksquare$
$\blacksquare\blacksquare\blacksquare\Box\blacksquare$
$\Box\Box\blacksquare\Box\blacksquare$
$\blacksquare\Box\blacksquare\Box\blacksquare$

Its last two columns have opposite colors, erase them both. The resulting rectangle has size $(n-2)\times(n-1)$ and it is the augmented rectangle one would consider to solve the $n-2$ case:

$\Box\Box\Box$
$\blacksquare\blacksquare\blacksquare$
$\Box\Box\blacksquare$
$\blacksquare\Box\blacksquare$

Either continue these erasure steps till the end (where one gets an evenly colored rectangle of size $1\times2$ if $n$ is odd and $2\times1$ if $n$ is even) or assume that the hypothesis holds at rank $n-2$, either way one is done.

Edit An alternative, maybe simpler, is to peel the $n\times(n+1)$ rectangle, starting from the top and right sides, one L-shape after the other. For example the first L-shape is the union of the $2n$ elementary squares in the last line and the last column.Then the result follows from the fact that these L-shapes are all evenly colored,