For the first one, $\displaystyle \sum_{k=1}^{n} k^2$, you can probably try this way.
$$k^2 = \binom{k}{1} + 2 \binom{k}{2}$$
This can be proved using combinatorial argument by looking at drawing $2$ balls from $k$ balls with replacement.
The total number of ways to do this is $k^2$.
The other way to count it is as follows. There are two possible options either you draw the same ball on both trials or you draw different balls on both trials. The number of ways for the first option is $\binom{k}{1}$ and the number of ways for the second option is $\binom{k}{2} \times \left( 2! \right)$
Hence, we have that $$k^2 = \binom{k}{1} + 2 \binom{k}{2}$$
$$\displaystyle\sum_{k=1}^{n} k^2 = \sum_{k=1}^{n} \binom{k}{1} + 2 \sum_{k=1}^{n} \binom{k}{2} $$
The standard combinatorial arguments for $\displaystyle\sum_{k=1}^{n} \binom{k}{1}$ and $\displaystyle\sum_{k=1}^{n} \binom{k}{2}$ gives us $\displaystyle \binom{n+1}{2}$ and $\displaystyle \binom{n+1}{3}$ respectively.
Hence, $$ \sum_{k=1}^{n} k^2 = \binom{n+1}{2} + 2 \binom{n+1}{3}$$
For the second case, it is much easier than the first case and in fact this suggests another method for the first case.
$k(k+1)$ is the total number of ways of drawing 2 balls from $k+1$ balls without replacement where the order is important. This is same as $\binom{k+1}{2} \times \left(2! \right)$
Hence, $$\sum_{k=1}^{n} k(k+1) = 2 \sum_{k=1}^{n} \binom{k+1}{2} = 2 \times \binom{n+2}{3}$$
This suggests a method for the previous problem since $k^2 = \binom{k+1}{2} \times \left(2! \right) - \binom{k}{1}$
(It is easy to give a combinatorial argument for this by looking at drawing two balls from $k+1$ balls without replacement but hide one of the balls during the first draw and add the ball during the second draw)
and hence $$\sum_{k=1}^{n} k^2 = 2 \times \binom{n+2}{3} - \binom{n+1}{2} $$
Best Answer
Ok, I think I've found an answer along with a proof. I will go through a series of claims (of course there may be more efficient ways of going about this). The main result is in Claim 3.
Claim 1. For nonnegative integers $m, r, t$ we have $$ \sum_{j=0}^{m+t} \binom{m+t}{j}\binom{j+r}{t+r}(-1)^{j} = (-1)^{m+t}\binom{r}{m}. \tag*{$(1)$} $$
Proof. We proceed by Egorychev's method. We will use the fact that $$ \binom{j+r}{t+r} = \frac{1}{2\pi i}\oint \frac{(1+z)^{j+r}}{z^{t+r+1}} \, dz, $$ where the integral is a contour integral over a circle $|z|=\varepsilon$ for some $0<\varepsilon<\infty$. Using this fact, and interchanging the summation and integral signs when necessary, we have \begin{align*} \sum_{j=0}^{m+t} \binom{m+t}{j}\binom{j+r}{t+r}(-1)^{j} &= \sum_{j=0}^{m+t} \binom{m+t}{j} \left( \frac{1}{2\pi i}\oint \frac{(1+z)^{j+r}}{z^{t+r+1}} \, dz \right) (-1)^{j} \\[1.2ex] &= \frac{1}{2\pi i}\oint \; \sum_{j=0}^{m+t} \binom{m+t}{j} \frac{(1+z)^{j+r}}{z^{t+r+1}} (-1)^{j} \, dz \\[1.2ex] &= \frac{1}{2\pi i}\oint \frac{(1+z)^{r}}{z^{t+r+1}} \; \sum_{j=0}^{m+t} \binom{m+t}{j} (1+z)^{j} (-1)^{j} \, dz \\[1.2ex] &= \frac{1}{2\pi i}\oint \frac{(1+z)^{r}}{z^{t+r+1}} \; \sum_{j=0}^{m+t} \binom{m+t}{j} (-1-z)^{j} \, dz \\[1.2ex] &= \frac{1}{2\pi i}\oint \frac{(1+z)^{r}}{z^{t+r+1}} (1 - 1 - z)^{m+t} \, dz \\[1.2ex] &= \frac{1}{2\pi i}\oint \frac{(1+z)^{r}}{z^{t+r+1}} (-z)^{m+t} \, dz \\[1.2ex] &= \frac{1}{2\pi i}\oint \frac{(1+z)^{r}}{z^{r-m+1}} (-1)^{m+t} \, dz \\[1.2ex] &= (-1)^{m+t}\binom{r}{r-m} \\[1.2ex] &= (-1)^{m+t}\binom{r}{m}. \end{align*} $$\tag*{$\blacksquare$}$$
Claim 2. For nonnegative integers $n, r, s$ with $s\ge r$, we have $$ \sum_{k=0}^{n} \binom{n-r}{k-r}\binom{k}{s}(-1)^{k} = (-1)^{n}\binom{r}{n-s}. \tag*{$(2)$} $$
Proof. If $s > n$, then both sides of $(2)$ are clearly $0$, so for the proof let us assume $s\le n$. Take $m = n-s$ and $t = s-r$ and plug these into $(1)$. We obtain $$ \sum_{j=0}^{n-r} \binom{n-r}{j}\binom{j+r}{s}(-1)^{j} = (-1)^{n-r}\binom{r}{n-s}. $$ We can shift the summation index by taking $j = k-r$ (where $k$ is the new index). By doing this and then multiplying both sides by $(-1)^{r}$, we get $(2)$. $$\tag*{$\blacksquare$}$$
Claim 3. For nonnegative integers $n, r, s$, we have $$ \boxed{ \sum_{k=0}^{n}\binom{n}{k}\binom{k}{r}\binom{k}{s}(-1)^{k} = (-1)^{n}\binom{n}{n-r,\, n-s,\, r+s-n} } $$ where the RHS involves the multinomial coefficient.
Proof. Without loss of generality, assume $s\ge r$. Consider the fact that $$ \binom{n}{k}\binom{k}{r} = \binom{n}{r}\binom{n-r}{k-r} $$ and use this to obtain $$ \sum_{k=0}^{n}\binom{n}{k}\binom{k}{r}\binom{k}{s}(-1)^{k} = \binom{n}{r}\sum_{k=0}^{n} \binom{n-r}{k-r}\binom{k}{s}(-1)^{k}. $$ By Claim 2, we see that the RHS reduces to $$ \binom{n}{r}\cdot (-1)^{n}\binom{r}{n-s}. $$ This is then decomposed as \begin{align*} \binom{n}{r}\cdot (-1)^{n}\binom{r}{n-s} &= (-1)^{n} \frac{n!}{r!(n-r)!}\frac{r!}{(n-s)!(r-n+s)!} \\ &= (-1)^{n}\frac{n!}{(n-r)!(n-s)!(r+s-n)!}, \end{align*} and the fraction in the last expression can be identified as a multinomial coefficient, giving us $$ \sum_{k=0}^{n}\binom{n}{k}\binom{k}{r}\binom{k}{s}(-1)^{k} = (-1)^{n}\binom{n}{n-r,\, n-s,\, r+s-n} $$ as desired. $$\tag*{$\blacksquare$}$$