Characterizations of Chromatic Polynomials Can somebody help me to prove these?

Let G be a nontrivial graph and $P(G,t)$ be the chromatic polynomial of G where $t$ is a nonnegative integer.

    
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*If $G$ is connected, then $t$ divides $P(G,t)$ but $t^2$ does not.
    
*If $G$ is a connected graph of order $n$, then there exists $n$ positive integers $a_1$, $a_2$, ... , $a_n$ such that $P(G,t)=a_nt^n$$-$$a_n$$_-$$_1$$t^n$$^-$$^1$$+$$a_n$$_-$$_2$$t^n$$^-$$^2$$...±$$a_1t$; i.e., the coefficients of $P(G,t)$ alternate in sign.
    
*If $P(G,t)=c_nt^n$$+$$c_n$$_-$$_1$$t^n$$^-$$^1$$+$$c_n$$_-$$_2$$t^n$$^-$$^2$$...+$$c_1t$, then the size of $G$ is |$c_n$$_-$$_1$|.
    
*The number of connected components of $G$ is equal to the lowest exponent of $t$ in $P(G,t)$.
    
  

I tried to use PMI to prove the first of the four statements. Yet, I still couldn't manage to finish each of them. For statement 1, I use induction on the size $m$ of $G$. For statements 2 and 3, I used induction on $n+m$ where $n$ is the order of $G$. I still can't find an elegant proof of statement 4. Can anybody help me? Your help is highly appreciated...
 A: First, the statement of Part $1$ is a little unclear, since $t=1$ is a nonnegative integer and both $1$ and $1^2$ certainly divide $P(G,1)$. I think what they are trying to say is that if we consider $P(G,t)$ as a polynomial in $t$, then the coefficient of $t$ is nonzero. 
Anyways, presumably you know the deletion/contraction formula $$P(G,t)=P(G-e,t)-P(G/e,t),$$ which holds for every edge $e\in E(G)$. One should also note that the constant coefficient of $P(G,t)$ is zero for every nontrivial $G$. (How many proper colorings of a nontrivial graph are there using zero colors?)
In addition to proving $1,2,3,4$, you should, at the same time, also prove:

$0.$ $P(G,t)$ is a monic polynomial of degree $n=|V(G)|$. 

So, yes, apply induction on $m$ for a connected graph $G$ with $m$ edges. I'll leave you to do the base case, and so let $e$ be an edge. Try to choose $e$ such that that $G-e$ is connected. If no such $e$ can be chosen, then $G$ is a tree and you can calculate $P(G,t)$ explicitly. Otherwise, by induction, we may write \begin{align*} P(G-e,t)&=t^n-a_{n-1}t^{n-1}+\cdots + (-1)^{n+1} a_1t\\ P(G/e,t)&=t^{n-1}-b_{n-2}t^{n-2}+\cdots + (-1)^{n} b_1t,\end{align*} for positive integers $a_1,\ldots,a_{n-1}$ and $b_1,\ldots,b_{n-2}$. Now, prove $0,1,2,3$ by plugging into the deletion-contraction formula. 
For $4$, use the fact (or prove it) that the chromatic polynomial of a graph is the product of the chromatic polynomials of its connected components and apply Part $1$. 
