Proof of a summation of $k^4$

Using the fact that $$\sum_{k=0}^{n}k^5=\sum_{k=0}^{n}(k+1)^5-(n+1)^5$$ Using Binomial Expansion $$\sum_{k=0}^{n}k^5=\sum_{k=0}^{n}k^5+5\sum_{k=0}^{n}k^4+10\sum_{k=0}^{n}k^3+10\sum_{k=0}^{n}k^2+5\sum_{k=0}^{n}k+\sum_{k=0}^{n}(1)-(n+1)^5$$ After Rearranging $$5\sum_{k=0}^{n}k^4=(n+1)^5 -10\sum_{k=0}^{n}k^3-10\sum_{k=0}^{n}k^2-5\sum_{k=0}^{n}k-\sum_{k=0}^{n}(1)$$ Replacing Known Sums $$5\sum_{k=0}^{n}k^4=(n+1)^5 -10\left(\frac{(n)(n+1)}{2}\right)^2-10\frac{(n)(n+1)(2n+1)}{6}-5\frac{(n)(n+1)}{2}-n$$ Then a bit of rearrangement and We get

$$ 30\sum_{k=1}^nk^4=n(n+1)(6n^3 + 9n^2 + n - 1)$$

$$ 30\sum_{k=1}^nk^4=n(n+1)(2n+1)(3n^2+3n-1)$$

$$\sum_{k=1}^{n}k^4=\sum_{k=0}^{n}k^4=\frac{(n)(n+1)(2n+1)(3n^2+3n-1)}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$$

Here's an alternative approach using binomial coefficients.

Firs we express $k^4$ as a linear combination of $\binom{k+a}{4}$ where $a=0,1,2,3$, i.e. $$k^4={k+3\choose 4}+11{k+2\choose 4}+11{k+1\choose 4}+{k\choose 4}$$.

Summing this from $1$ to $n$ and using the hockey stick summation identity, i.e. $\sum_{r=0}^m {r\choose b}={m+1\choose {b+1}}$, we have $$\begin{align} \sum_{k=1}^nk^4&=\sum_{k=1}^n\left[{k+3\choose 4}+11{k+2\choose 4}+11{k+1\choose 4}+{k\choose 4}\right]\\ &={n+4\choose 5}+11{n+3\choose 5}+11{n+2\choose 5}+{n+1\choose 5}\\ \end{align}$$

which is convenient for numerical evaluation.

This reduces to $$\begin{align} &\frac{(n+4)^\underline{4}+11(n+3)^\underline{4}+11(n+2)^\underline{4}+(n+1)^\underline{4}}{5!}\\ &=\frac 1{5!}n(n+1)\biggr[(n+4)(n+3)(n+2)+(n-1)(n-2)(n-3)+11(n+2)(n-1)\bigg((n+3)+(n-2)\biggr)\biggr]\\ &=\frac 1{30}n(n+1)(2n+1)(3n^3+3n-1) \end{align}$$

where $x^\underline{n}$ represents the falling factorial and is given by $\underbrace{x(x-1)(x-2)\cdots(x-n+1)}_{n\text{ terms}}$.