generating function formula Suppose $\lambda$  is a young tableau, with one color boxes. 
Prove the generating function, 
$F(x) = \Sigma_{\lambda} x^{i} = \prod_{n \geq 1} \frac{1}{1-x^n} $, where $i$ is the number of boxes in the young tableau. 
attempt: 
$\Sigma_{\lambda} x^{i} = \prod_{n \geq 1} \frac{1}{1-x^n} =  \frac{1}{1-x} \cdot \frac{1}{1-x^2} \cdot \frac{1}{1-x^3} \cdot \frac{1}{1-x^4}\cdots = (1 + x + x^2 + x^3 + \cdots)(1 + x^2 + x^4+ \cdots )(1 + x^3 + x^6 + x^9 + \cdots)( 1+ x^4 + x^8 + x^{12} \cdots ) = 1 + x + 2x^2 + 3x^3 + \cdots$ .
Here this one comes from when we have zero boxes so we would have $x^0$, and when there is only one box, then we have $x^1 $, similarly, there are $2x^2$ since there are two ways of having a young tableau with two boxes (horizontally, and vertically), and so on.
Can someone please help me?I am stuck after this. I would really appreciate it.
Thank you.

Let $p(n,k)$ denote the number of partitions of $n$ with largest part at most $k$. Then, of the $p(n,k)$ such partitions, there are some that have a partition with part $k$ and some that don't. Those that don't have part $k$ are all precisely the partitions of $n$ with largest part at most $k-1$, so there are $p(n,k-1)$ of them. Those that do have part $k$ can be converted into a partition of largest part $k$ of $n-k$ by removing the $k$. Thus, we have the recurrence
$$
p(n,k) = p(n-k,k) + p(n,k-1). 
$$
Defining the generating function $F_k(x) = \sum_{n\geq0}p(n,k)x^n$, the recurrence translates to
$$
F_k(x) = x^kF_k(x) + F_{k-1}(x)
$$
which rearranges to
$$
F_k(x) = \frac{F_{k-1}(x)}{1-x^k}. 
$$
Can you proceed from here?
Edit:
Now, note that $F_1(x) = \sum_{n\geq0} x^n = \frac1{1-x}$ since there is always exactly one way to write $n$ as a partition with largest part $1$. Then, induction shows that $F_k(x) = \prod_{1\leq n\leq k}\frac1{1-x^n}$. It follows that for partitions of any large sized part, you take the desired infinite product. 
