Show by induction on $n$ that: $$1^4 + 2^4 +\cdots+n^4=\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30}$$
I proved true for case $n=1$ , assumed true for $n=k$ , but cannot get things to work out. I tried putting the right side over a common denominator and going from there but I'm sure how much it helps.
 A: $$\sum_{k=1}^{n+1}k^{4}=\sum_{k=1}^{n}k^{4}+\left(n+1\right)^{4}=\frac{n^{5}}{5}+\frac{n^{4}}{2}+\frac{n^{3}}{3}-\frac{n}{30}+n^{4}+4n^{3}+6n^{2}+4n+1
 $$$$=\frac{6n^{5}+15n^{4}+10n^{3}-n}{30}+\frac{30n^{4}+120n^{3}+180n^{2}+120n+30}{30}=\frac{6n^{5}+45n^{4}+130n^{3}+180n^{2}+119n+30}{30}$$
 and see that$$6\left(n+1\right)^{5}+15\left(n+1\right)^{4}+10\left(n+1\right)^{3}-\left(n+1\right)$$$$=\left(6n^{5}+30n^{4}+60n^{3}+60n^{2}+30n+6\right)
 +\left(15n^{4}+60n^{3}+90n^{2}+60n+15\right)+\left(10n^{3}+30n^{2}+30n+10\right)-n-1=6n^{5}+45n^{4}+130n^{3}+180n^{2}+119n+30.$$
A: First, show that this is true for $n=1$:
$\sum\limits_{k=1}^{1}k^4=\frac{6\cdot1^5+15\cdot1^4+10\cdot1^3-1}{30}$
Second, assume that this is true for $n$:
$\sum\limits_{k=1}^{n}k^4=\frac{6n^5+15n^4+10n^3-n}{30}$
Third, prove that this is true for $n+1$:
$\sum\limits_{k=1}^{n+1}k^4=$
$\color{red}{\sum\limits_{k=1}^{n}k^4}+(n+1)^4=$
$\color{red}{\frac{6n^5+15n^4+10n^3-n}{30}}+(n+1)^4=$
$\frac{6n^5+15n^4+10n^3-n}{30}+n^4+4n^3+6n^2+4n+1=$
$\frac{6n^5+15n^4+10n^3-n}{30}+\frac{30n^4+120n^3+180n^2+120n+30}{30}=$
$\frac{6n^5+45n^4+130n^3+180n^2+119n+30}{30}=$
$\frac{6(n+1)^5+15(n+1)^4+10(n+1)^3-(n+1)}{30}$

Please note that the assumption is used only in the part marked red.
