Is it normal to have this much difficulty with the problems in Spivak's Calculus? So I'm a senior in high school going through Spivak's Calculus. I've made it up to chapter 3 with a lot of difficulty and only being able to solve less than half of the problems. If I can't prove some theorem on my own I can usually understand the proof although sometimes it takes whole days to feel like I'm starting to wrap my head around it. My biggest problem is that it seems like I never know where to begin. Is this something that will just come with time? For example a problem I'm currently stuck on is showing that for each $n$ there is a polynomial function of degree $n$ with $n$ roots. I have already proven that if $f$ is a polynomial function of degree $n$ , then $f$ has at most $n$ roots. I assume this proof builds on that one. I just don't know where to start. 
Update: Okay so funny enough I think a proof just came to me but I want to make sure it's correct. So my proof is

There is a polynomial of degree $n=1$ with exactly one root. Assume $g$ is a polynomial of degree $k$ with $k$ roots, then $f(x)=(x-a)g(x)$ is a polynomial of degree $k+1$ with $k+1$ roots. Therefore for each $n$ there is a polynomial function of degree $n$ with $n$ roots. 

Is this proof valid?
 A: My solution
With just implicit induction, you could simply take the polynomial $f(x)=(x-1)(x-2)\dotsm(x-n)$. If you have already proved that the degree of a product is the sum of the degrees and that $r$ is a root of $p(x)q(x)$ if and only if it is either a root of $p$ or a root of $q$, then you're done.
Your attempt
It depends on what you mean by “$n$ roots”. I believe that Spivak means $n$ distinct roots, otherwise simply $x^n$ would be an answer.
Your idea is good, but lacks a detail. OK, the polynomial $x$ has degree $1$ and $1$ root. Suppose that we're able to find a polynomial $g$ of degree $n$, having $n$ (distinct) roots $a_1,a_2,\dots, a_n$. Then take $a_{n+1}=\max\{a_1,\dots,a_n\}+1$ and the polynomial $f(x)=(x-a_{n+1})g(x)$ has degree $n+1$ and $n+1$ distinct roots.
Comment
The strange choice of $a_{n+1}$ ensures it is distinct from the given roots. In other contexts, the statement would be false: over the two-element field, for instance, no polynomial of degree $n>2$ can have $n$ distinct roots.
A: Update: Okay so funny enough i think a proof just came to me but i want to make sure it's correct. So my proof is There is a polynomial of degree $n=1$ with exactly one root. Assume $g$ is a polynomial of degree $k$ with $k$ roots, then $f(x)=(x-a)g(x)$ is a polynomial of degree $k+1$ with $k+1$ roots. Therefore for each $n$ there is a polynomial function of degree $n$ with $n$ roots. Is this proof valid?
