This comes from I. G. MacDonald's book Symmetric Functions and Hall Polynomials:
(1.7'): Put $f_{\lambda,n}=\sum_{j=1}^{n}t^{\lambda_j+n-j}$, then $$ f_{\lambda,n}(t)+t^{m+n-1}f_{\lambda',m}(t^{-1})=(1-t^{m+n})/(1-t) $$ where $m \geq \lambda_1$, $n \geq \lambda_1'$. If $\lambda=(\lambda_1,\cdots)$ is a partition, then $\lambda'=(\lambda'_1,\cdots)$ is its conjugate where $\lambda'_i=|\{j:\lambda_j \geq i\}|$
I'm OK with how he derived (2) but why does (2) also implie (3)? For him it seemed to be so trivial but I guess not for everyone, although I think the calculation shouldn't be tedious.
Equation (2) is such that the multiset of terms for each sum on the left side have a matching term on the right side. All of those terms are transformed by $t^\square \mapsto (1 - t^\square)$, so that the matching of left-side terms and right-side terms remains. The equation
$$\prod_{x \in \lambda} \left(1 - t^{h(x)}\right) \cdot \prod_{i < j} \left(1 - t^{\mu_i - \mu_j}\right) = \prod_{i \geq 1} \prod_{j=1}^{\mu_i} \left(1 - t^j\right)$$
follows by matching the factors on the left side with the right in the same manner as the sums in (2).