Mathematical induction in series How can you prove by mathematical induction that, for all positive integers of $n$ , $p(p+1)$ is a factor of the monic polynomial of $p$ in the sum of
$$
1^n + 2^n +3^n + \dotsb + p^n 
$$
I have got this issue when comparing the situation of $n = 1, 2 ,3$ and few more initial stages because sum of those cases can be found by other means. But I don't know how to connect assumption to the next stage of the proof. Your suggestions are greatly appreciated. 
When you consider $n=1$ the sum is $1+2+....+p$ and since this is arithmetic series answer is $\frac{p(p+1)}{2}.$ Here the monic polynomial is $p(p+1)$ therefore the result is true for $n=1.$
If you are interested you can further investigate this in the cases $n=2,3,4..$and it works. But main issue is what is the relationship between two consecutive cases, I mean such as between $n=1$ and $n=2.$ If there is a such relationship applying mathematical induction may be not difficult to this issue.
Can we get an idea by the help of Faulhaber's polynomials?
I have raised this issue because of my curiosity related to the pattern of Faulhaber's polynomials and I have come across many results in connection with these polynomials but couldn't able to connect with the mathematical induction. I doubt sometimes there may be a way by assuming for $ n = 1,2,...k $, and trying to prove for $ n= k+1 $
When you study the history of mathematics many concepts were formulated because of the skill of observing patterns. Therefore I think you can realize the importance of questioning this kind of issues .
 A: We can prove that the $n^\text{th}$ Faulhaber polynomial contains $p(p+1)$ as a factor, for all $n\ge 1$. To do this, we use the following formula which converts normal powers to falling factorial powers.
$$k^n = \sum_{i=1}^n{n \brace i}k_{(i)}, \qquad (n\ge 1)$$
where $k_{(i)}=k(k-1)\cdots(k-i+1)$ is the falling factorial, and where ${n \brace i}$ is a Stirling number of the second kind.
The benefit of introducing falling factorials is that they are summed easily:
$$
\sum_{k=0}^p k_{(i)}=\frac{(p+1)_{(i+1)}}{i+1}
$$
Therefore, we use this identity to find the Faulhaber polynomial:
$$
\sum_{k=0}^p k^n
=\sum_{k=0}^p\sum_{i=1}^n{n\brace i}k_{(i)}
=\sum_{i=1}^n\sum_{k=0}^p{n\brace i}k_{(i)}
=\sum_{i=1}^n{n \brace i}\frac{(p+1)_{(i+1)}}{i+1}
$$
The upshot is that the final polynomaial is a linear combination of polynomials of the form
$$
(p+1)_{(i+1)}=(p+1)\cdot p\cdots (p-i+1).
$$
In particular, all of these polynomial summands contain $p(p+1)$ as a factor, so the total polynomial does as well.
