As mentioned in the comment above, generating functions are the standard tool for proving such statements. If you can write down the proper generating functions, proofs sometimes just fall right out.
Let $S = \{ \ell^2 \;|\; \ell \in \mathbb{N} \}$ be the set of perfect squares.
In the following generating function: the coefficient of $x^k$ is the number of partitions of $k$ which do not involve $d$ or more copies of $d$ for each $d$.
$ \prod\limits_{d=1}^\infty \left(\sum\limits_{j=0}^{d-1} x^{jd}\right) =$
$(1+x^2)(1+x^3+x^6)(1+x^4+x^8+x^{12})\cdots (1+x^d+x^{2d}+\cdots+x^{d(d-1)}) \cdots $
$=((1-x^2)^{-1}-x^{2^2}(1-x^2)^{-1})((1-x^3)^{-1}-x^{3^2}(1-x^3)^{-1})\cdots ((1-x^k)^{-1}-x^{k^2}(1-x^k)^{-1}) \cdots $
$=\left(\frac{1-x^{1^2}}{1-x^1}\right)\left(\frac{1-x^{2^2}}{1-x^2}\right)\left(\frac{1-x^{3^2}}{1-x^3}\right)\cdots \left(\frac{1-x^{k^2}}{1-x^k}\right) \cdots $
$=\prod\limits_{k = 1}^\infty \frac{1-x^{k^2}}{1-x^k} = \prod\limits_{k = 1}^\infty \frac{\frac{1}{1-x^k}}{\frac{1}{1-x^{k^2}}} =\frac{\prod\limits_{k = 1}^\infty \frac{1}{1-x^k}}{\prod\limits_{k=1}^\infty\frac{1}{1-x^{k^2}}} = \prod\limits_{k\in\mathbb{N}-S} \frac{1}{1-x^k}$
In the final generating function: the coefficient of $x^k$ counts the number of partitions not involving perfect squares.
Edit For a bit more about generating functions here is a link to another question I answered: partitions and generating functions