Prove formula from Pascal's triangle I found this in my Norwegian mathematics book. I have solved a)-d) but am stuck at e). Nice if you have feedback on what I have attempted to do.
a) Where are the numbers of the form $\binom{k}{l}$ in Pascal's triangle? Where are the numbers of the form $\binom{k}{2}$? In general, if i is a fixed number and k varies, where will you find the numbers $\binom{k}{i}$?
My answer: I find it on the third diagonal line from the left. If i is fixed, you find it on the ith diagonal line from the left
b) Check through some examples that the following equality seems to hold:
$\sum_{k=1}^n \binom{k}{i} = \binom {n+1}{i+1}$
Can you explain it through Pascals triangle?
My answer: 
$\sum_{k=2}^4\binom{k}{2}=\binom{5}{3}=10$
In Pascal's triangle this is the sum all from the third diagonal line from the left up to k=4.
The result is $\binom {n+1}{i+1}$
c) Prove the formula b) by induction on n.
My answer:
The equation holds for n. Assuming that the equation also holds for n+1.
$\sum_ {k=1}^{n+1}\binom{n+1}{i+1}+\binom{n+1}{i}$
$=\frac{(n+1)!}{(i+1)!(n+1)!}+\frac{(n+1)!}{i!(n+1-i)!}$
$=\frac{(n+1)!(n+1-i)}!{(i+1)!(n+1-i)!}+\frac{(n+1)!(i+1)}{(i+1)!(n+1-i)!}$
$=\frac{(n+1)!(n+1-i+i+1)}{(i+1)!(n+1-i)!}$
$=\frac{(n+2)!}{(i+1)!((n+2)-(i+1))!}$
$=\binom{n+2}{i+1}$
This is correct according to Pascal's triangle.
d) Show that the formula can be rewritten to:
$\sum_{k=1}^n k(k-1)(k-2)\dots (k-i+1)=\frac{(n+1)n(n-1)\dots(n-i+1)}{i+1}$
My thoughts: Multiply by (n-i)! / (k-1)! on both sides.
$\sum_{k=1}^n \binom{k}{i} = \binom {n+1}{i+1}$
$\sum_{k=i}^n\frac{k!}{i!(k-i)!}\dot (k-i)!=\sum_{k=1}^nk(k-1)(k-2)\dots(k-i+1)$
$\binom {n+1}{i+1}\dot (n-i)!=\frac{(n+1)!}{(i+1)!(n-i)!}\dot (n-1)!$
$=\frac{(n+1)n(n-1)\dots(n-i+1)}{i+1}$
e) Show how the formula in d) can be used to find expressions for the sums $\sum_{k=1}^nk^2$, $\sum_{k=1}^nk^3$ etc. Hint: Start with the formula
$\sum_{k=1}^nk=\frac{n(n+1)}{2}$
And work your way up the potentials.
My thoughts: I just know how to get the formula $\sum_{k=1}^nk=\frac{n(n+1)}{2}$

I have the solution for $\sum_{k=1}^nk^2$
$k^2=k(k-1)+k$ so that $\sum_{k=1}^nk^2=\sum_{k=1}^nk(k-1)+\sum_{k=1}^nk$
$\sum_{k=1}^n k(k-1)(k-2)\dots (k-i+1)=\frac{(n+1)n(n-1)\dots(n-i+1)}{i+1}$
$\sum_{k=1}^n k(k-1)=\frac{(n+1)n}{2+1}=\frac{(n+1)n(n-1)}{3}$
$\sum_{k=1}^nk^2=\sum_{k=1}^nk(k-1)+\frac{n(n+1)}{2}=\frac{(n+1)n(n-1)}{3}+\frac{n(n+1)}{2}=\frac{2(n+1)n(n-1)}{3*2}+\frac{3n(n+1)}{2*3}=\frac{n(n+1)(2(n-1)+3))}6=\frac{n(n+1)(2n+1)}6$
The solution for $\sum_{k=1}^nk^3$
$\sum_{k=1}^n k(k-1)(k-2))=\sum_{k=1}^nk^3-\sum_{k=1}^n3k^2+2\sum_{k=1}^n3k=\frac{(n+1)n(n-1)(n-2)}{4}$
$\sum_{k=1}^nk^3 -3\frac{n(n+1)(2n+1)}6 + 2\frac{n(n+1)}{2}=\frac{(n+1)n(n-1)(n-2)}{4}$
$\sum_{k=1}^nk^3=\frac{(n+1)n(n-1)(n-2)}{4}+3\frac{n(n+1)(2n+1)}6-2\frac{n(n+1)}{2}$
$\frac{2(n+1)n(n-1)(n-2)}{8}+4\frac{n(n+1)(2n+1)}8-8\frac{n(n+1)}{8}$
$=\frac{n^2(n+1)^2}{4}$
