(i).Prove that $\pi_m(n)=\pi_m(n-m)+\pi_{m-1}(n)$ without using the generating functions for $\pi_m(n)$. Questions:
$\pi_m(n)$ is defined as the number of partitions of n in which each part is no larger than m.
(i).Prove that $\pi_m(n)=\pi_m(n-m)+\pi_{m-1}(n)$ without using the generating functions for $\pi_m(n)$.

The generating functions for $\pi_m(n)$ is $\sum\limits_{n=0}^\infty\pi_m(n)q^n=\prod\limits_{j=1}^m \frac{1}{1-q^j}$, where $|q|<1$.

(ii). Deduce from (i) that if $F_m(q)=\sum\limits_{n=0}^\infty\pi_m(n)q^n$, then $$F_m(q)=q^mF_m(q)+F_{m-1}(q).$$
(iii) Use (ii) to prove the generating functions for $\pi_m(n)$.
My works:
I really have no idea how to prove (i);
(ii)
$F_{m-1}(q)=\sum\limits_{n=0}^\infty\pi_{m-1}(n)q^n$;
I'm not sure if these steps are correct:
$\sum\limits_{n=0}^\infty\pi_m(n-m)q^{n}
=q^{n}(\pi_m(0-m)+\pi_m(1-m)+...+\pi_m(m-m)+\pi_m(m+1-m)+...+\pi_m(\infty-m))
=q^{m}(\pi_m(0-m)+\pi_m(1-m)+...+\pi_m(m-1-m))+q^{n}(\pi_m(m-m)+\pi_m(m+1-m)+...+\pi_m(\infty-m))
=q^{n+m}(0+...+0+\pi_m(0)+\pi_m(1)+...+\pi_m(\infty))
=q^{n+m}(\pi_m(0)+\pi_m(1)+...+\pi_m(\infty))
=q^{n+m}\sum\limits_{n=0}^\infty\pi_m(n)=q^m F_m(q)$
Then,
$$F_m(q)=\sum\limits_{n=0}^\infty\pi_m(n)q^n=\sum\limits_{n=0}^\infty(\pi_m(n-m)+\pi_{m-1}(n))q^n=\sum\limits_{n=0}^\infty\pi_m(n-m)q^n+\sum\limits_{n=0}^\infty\pi_{m-1}(n)q^n=q^m F_m(q)+F_{m-1}(q)$$
 as desire.
(iii)
We need to prove $F_m(q)=\prod\limits_{j=1}^m \frac{1}{1-q^j}$, where $|q|<1$.
$\prod\limits_{j=1}^m \frac{1}{1-q^j}=\frac{1}{1-q}\frac{1}{1-q^2}\frac{1}{1-q^3}...\frac{1}{1-q^m}=(1+q+q^2+q^3+...)(1+q^2+q^4+...)(1+q^3+q^9+...)...(1+q^n+q^{2n}+...)=1+q+(q^2+q^2)+(q^3+q^3+q^3)+...=1+1*q^1+2*q^2+3*q^3+...=\pi_m(0)+\sum\limits_{n=1}^\infty nq^n$
Does $\pi_m(0)+\sum\limits_{n=1}^\infty nq^n=\sum\limits_{n=0}^\infty\pi_m(n)q^n$?
I think I am suppose to use (ii) to prove it...but I have no idea how $q^mF_m(q)+F_{m-1}(q)=\prod\limits_{j=1}^m \frac{1}{1-q^j}.$

After reading the Hints given by Mike:
(iii) By (ii), we know $(1-q^m)F_m(q)=F_{m-1}(q) \Rightarrow \frac{F_m(q)}{F_{m-1}(q)}=\frac{1}{1-q^m}$
$\prod\limits_{j=1}^m \frac{1}{1-q^j}=\frac{1}{1-q^1}\frac{1}{1-q^2}...\frac{1}{1-q^m}=\frac{F_1(q)}{F_{0}(q)}\frac{F_2(q)}{F_{1}(q)}...\frac{F_m(q)}{F_{m-1}(q)}=\frac{F_m(q)}{F_0(q)}=\sum\limits_{n=0}^\infty\pi_m(n)q^n$
  as desire.

 A: By abuse of notation, I'm also going to denote $\pi_m(n)$ also as the set of partitions of $n$ into parts at most $m$.
Hint for (i): Seperate $\pi_m(n)$ into two sets: those partitions that have $m$ as a part, and those partitions whose parts are all strictly less than $m$. The two sets on the right are the partitions of $n-m$ into parts at most $m$ and the partitions of $n$ into parts at most $m-1$. Do you see a bijection between each pair of sets?
(ii) You're working too hard (and in your attempted proof, you added infinitely many positive numbers, so your function would diverge). $$F_m(q)=\sum_{n=0}^\infty \pi_m(n) q^n = \sum_{n=0}^\infty \pi_{m-1}(n)q^{n} + \sum_{n=0}^\infty \pi_m(n-m)q^{n}$$ As you noted, the first $m$ terms vanish from the second sum. Re-indexing, we have $$F_m(q) = \sum_{n=0}^\infty \pi_{m-1}(n)q^n + \sum_{n=0}^\infty \pi_m(n)q^{n+m} = F_{m-1}(q) + q^m\sum_{n=0}^\infty \pi_m(n)q^{n}$$ $$= F_{m-1}(q) + q^mF_m(q)$$
Hint for (iii) Use induction, and rewrite (ii) as $(1-q^m)F_m(q) = F_{m-1}(q)$
