Prove that if $0\leq a_k \leq b_k$ then $\prod_{k=1}^n a_k \leq \prod_{k=1}^{n} b_k$ Let $0\leq a_k\leq b_k$ for all $k\in\mathbb{N}$, then show
$$\prod_{k=1}^{n} a_k \leq \prod_{k=1}^{n} b_k$$
My attempt: since $0\leq a_k\leq b_k$ for all $k\in\mathbb{N}$ we have that 
$$0 \leq a_0\leq b_0, 0 \leq a_1 \leq b_1,..., 0\leq a_k \leq b_k$$
Multiplying for $a_1$ the first inequality we have
$$0\leq a_0a_1 \leq b_0a_1$$
But $a_1 \leq b_1$, so we have
$$0\leq a_0a_1 \leq b_0a_1 \leq b_0b_1$$
Doing this $n$ times we have
$$\prod_{k=1}^{n} a_k \leq \prod_{k=1}^{n} b_k$$
Two questions:
(1) Is this correct? I've done this by myself and even if this seems obvious I want to be sure or learn for my mistakes;
(2) If I want to remove the hypothesis $a_k \geq 0$ and $b_k \geq 0$, should I assume $|a_k|\leq|b_k|$ for all $k\in\mathbb{N}$ and prove
$$\prod_{k=1}^{n} |a_k| \leq \prod_{k=1}^{n} |b_k|$$
Thanks for your time.

1: Yes, what you've done is correct.
2: This follows immediately from the first part you have proven: Just define $\hat{a}_k = \left|a_k\right|$ and $\hat{b}_k = \left|b_k\right|$, then $\hat{a}_k$ and $\hat{b}_k$ satisfy the assumptions from your first point and you get
$$\prod_{k=1}^{n} \hat{a}_k \leq \prod_{k=1}^{n} \hat{b}_k$$ and by inserting the definition of $\hat{a}_k$ and $\hat{b}_k$ you get
$$\prod_{k=1}^{n} \left|a_k\right| \leq \prod_{k=1}^{n} \left|b_k\right|$$
