Q:

Proving the set of polynomial function is a monoid of functions

A function $f:\mathbb{R}\rightarrow \mathbb{R}$ is said to be a $polynomial function$ if there is a natural number $n$ and real numbers $f_0,f_1,\dots,f_n$ such that $$f(x)=f_nx^n+\dots f_1x+f_0$$ for x in $\mathbb{R}$.  Show that the set of all polynomial functions forms a monoid of functions on $\mathbb{R}$.
My attempt: So im new to abstract algebra and these type of proofs, so I apologize if this is a very easy question.   I know that we need to show that it is closed under composition and there  exist's an identity element.   I'm unsure how to go about that, I'm only necessarily looking for a hint to get me started, however full awnsers will be accepted too! 

A:

The polynomial functions under composition form a mnoid. 
Notice that the identity function is a polynomial, it is $f(x)=x$.
Now we must only prove that if $f$ and $g$ are polynomials then $f\circ g$ is a polynomial. 
Let $f(x)=a_nx^n+\dots+a_0$ and $g(x)=b_mx^m+\dots+b_0$.
Notice that $f\circ g(c)=f(b_mx^m+\dots+b_0)=a_n(b_mx^m+\dots+b_0)^n+a_{n-1}(b_mx^m+\dots+b_0)^{n-1}+\dots+ a_0$.
You must prove that each summand is a polynomial then you would be done ( prove that the sum of polynomials is a polynomial). 
The reason why each summand is a polynomial is that the product of polynomials is also a polynomial (you should also prove this).