$Q$ is a function from $\Bbb N$ to itself, $Q(n)-Q(n-1)=T(n)$ where $T$ is some polynomial of degree $k$, prove $Q$ is a polynomial of degree $k+1$. 
$Q$ is a function from $\mathbb N$ to itself, $Q(n)-Q(n-1)=T(n)$ where $T$ is some polynomial of degree $k$. Prove that $Q$ is a polynomial of degree $k+1$.

I've been given the above problem in an "elementary mathematics" course problem set. I've shown that $Q(n)=T(n)+T(n-1)+...T(1)+Q(0)$ (I'm including $0$ in the naturals) but that doesn't seem to clarify things. I know I can construct a polynomial $P$ of degree $k+1$ such that $P(n)=Q(n)$ for $0\leq n \leq k+1$ but I couldn't show that it is equal for all other numbers. Other than those attempts I don't know how to proceed, I can't think up other ways to show that a function is a polynomial other than explicitly constructing one that fits.
Help would be appreciated.
 A: A simple way to think about this problem of polynomial summation
is to use binomial coefficients. Consider the sequence of polynomial
functions
$\,C_m(n) := \binom{n}{m} \,$ which map $\,\mathbb{N}\,$ to $\,\mathbb{N}\,$.
They have the properties that
$\,C_m(n)-C_m(n-1) = C_{m-1}(n-1),\,$
$\,C_m(n)\,$ is of degree $\,m,\,$ and
$\,C_m(n) = 0\,$ if $\,0\le n<m.\,$
Notice that given any polynomial function
$\,f: \mathbb{N}\to\mathbb{N},\,$ the properties can be used to
prove that $\,f(n)\,$ can be expressed as a sum of binomial
coefficients and thus to express the partial sums of
$\,f(n)\,$ as a corresponding sum of binomial coefficients.
For example, suppose $\,f(n) = \sum_{m\in\mathbb{N}} a_m\,C_m(n).\,$
Then $\,a_0 = f(0) \in\mathbb{N}\,$ using the $3$rd property. Now,
similarly we can show that $\,a_1 = f(1)-f(0)\in\mathbb{Z}.\,$ In
general, the coefficients
$\,a_m = \Delta^mf(n)|_{n=0}\in\mathbb{Z}\,$
are the entries in the difference table of $\,f(n)\,$ at $\,n=0.$
For your situation, let $\,f(n)=Q(n)-Q(n-1)=T(n)\,$ where $\,T\,$
is of degree $\,k\,$ which implies that
$\,Q(n)= Q(0) + \sum_{m=0}^k a_{m+1}\,C_{m+1}(n+1)\,$
which is of degree $\,k+1.$
