Proof the Existence and Uniqueness of Factorization Form of Polynomial with Complex Coefficient If $p\in P(\Bbb{C})$ is a nonconstant polynomial, then $p$ has a
  unique factorization (except for the order of the factors) of the form
  
  $$p(z)=c(z-\lambda_1)....(z-\lambda_m)$$
  
  where $c,\lambda_1,....\lambda_m \in \Bbb{C}$


I know how to prove the existence of the form. However, I don't know how to do it with the uniqueness of the form. 

I know that $c$ must be unique, because $c$ equals the coefficient of $z^m$ in $p$. But how to prove that each $\lambda_j$ is unique? 

And I also don't understand why the coefficient must be $\Bbb{C}$, but why doesn't this apply to $\Bbb{R}$?
You can show that in $\mathbb{C}$, all linear functions are prime, and there do not exist primes which are not linear. 

Lemma 1: If $p$ is prime and $p$ divides $ab$, then $p$ divides $a$ or $p$ divides $b$. Let $p$ not divide $a$. Then as $p$ is relatively prime to $a$, $gcd(p,a) = 1$ and so $\exists x,y \in F$ such that $xp + ya = 1$. Multiplying by $b$, we have that $bxp + yab = b$. And we know that $p$ divides $ab$ so $ab = mp$. Thus we have that $bxp + ymp = b$ and so $p(bx + ym) = b$ so $p$ divides b. 

The Result: Now, let $f(x) = p_1 \cdot p_2 \cdot p_3 ... \cdot p_n$ where $p_1,.... p_n$ are prime factors. Suppose $f(x) = q_1 \cdot q_2 \cdot ... q_k$ where all $q_i$ are prime factors [We are trying a different factorisation]. Then, each $q_i$ must divide $f(x) = p_1 \cdot p_2 \cdot p_3 ... \cdot p_n$ and by Lemma 1, they must divide some $p_i$. But a prime is such that if $q_i$ divides $p_i$ then $q_i$ is an associate of $p_j$.  If we ignore the difference between a function and its associate, we can see that $q_i = p_j$ [effectively] and so we must have that each $q_i$ corresponds to a different $p_j$ and so we have that $n = k$ and the factorisation is unique.

Note: In $\mathbb{R}$, there can be quadratic equations which are prime as pointed out in the comment (They have complex factors only). Therein lies a fundamental difference.
You can show that in $\mathbb{C}$, all linear functions are prime, and there do not exist primes which are not linear. 

Lemma 1: If $p$ is prime and $p$ divides $ab$, then $p$ divides $a$ or $p$ divides $b$. Let $p$ not divide $a$. Then as $p$ is relatively prime to $a$, $gcd(p,a) = 1$ and so $\exists x,y \in F$ such that $xp + ya = 1$. Multiplying by $b$, we have that $bxp + yab = b$. And we know that $p$ divides $ab$ so $ab = mp$. Thus we have that $bxp + ymp = b$ and so $p(bx + ym) = b$ so $p$ divides b. 

The Result: Now, let $f(x) = p_1 \cdot p_2 \cdot p_3 ... \cdot p_n$ where $p_1,.... p_n$ are prime factors. Suppose $f(x) = q_1 \cdot q_2 \cdot ... q_k$ where all $q_i$ are prime factors [We are trying a different factorisation]. Then, each $q_i$ must divide $f(x) = p_1 \cdot p_2 \cdot p_3 ... \cdot p_n$ and by Lemma 1, they must divide some $p_i$. But a prime is such that if $q_i$ divides $p_i$ then $q_i$ is an associate of $p_j$.  If we ignore the difference between a function and its associate, we can see that $q_i = p_j$ [effectively] and so we must have that each $q_i$ corresponds to a different $p_j$ and so we have that $n = k$ and the factorisation is unique.

Note: In $\mathbb{R}$, there can be quadratic equations which are prime as pointed out in the comment (They have complex factors only). Therein lies a fundamental difference.