Use a generating series to show that every nonnegative integer $n$ can be written uniquely in the form $\sum_{k\geq 0} a_k k!, 0\leq a_k\leq k, k\geq 0$.
I'm not sure how to find a generating series. I know that $x(1-x)^{-2} = x(1+x+x^2+\cdots)(1+x+x^2+\cdots) = x+2x^2+3x^3 + \cdots.$ However, I can prove this by induction as follows. For the base case, $0 = 0\cdot 0!,$ which is the unique representation for $0$. Now assume that for all $n<j!, j\geq 1$ that $n$ can be written in the form $\sum_{k = 0}^{j-1} a_{k} k!.$ Now let $n'$ be so that $j! \leq n' < (j+1)!.$ Define $a:= \max \{b | n' - b\cdot j! \geq 0\}.$ Then $n ' - a\cdot j! := n_0$ satisfies that $0\leq n_0 < j!.$ Hence, by the induction hypothesis, $n_0 = \sum_{k = 0}^{j-1} a_k k!\Rightarrow n' = \sum_{k=0}^j a_kk!, a_{j} := a.$
One way to approach this problem is with induction, but it can be done with generating functions alone, which these hints will show.
Hints
In general, given subsets $S_1,S_2,S_3,\dots$ of $\mathbb N$, the number of solutions to the constrained equation $$ x_1+x_2+x_3+\dots=n,\\ \text{ such that } x_i\in S_i\text{ for all }i\in \{1,2,\dots\} $$ is the coefficient of $x^n$ in the generating function $$ \left(\sum_{i\in S_1}x^i\right)\left(\sum_{i\in S_2}x^i\right)\left(\sum_{i\in S_3}x^i\right)\cdots $$ For your problem, you want to show that every integer is representable in at least one way as a sum $$ a_0\cdot 0!+a_1\cdot 1!+a_2\cdot 2!+a_3\cdot 3!+\dots $$ where $a_i\in \{0,1,\dots,k\}$. Equivalently, letting $x_k=a_k\cdot k!$, this is equivalent to showing these is at least one solution to $x_1+x_2+\dots=n$, with the constraint that $$ x_k\in \{0,k!,2k!,\dots,k\cdot k!\} $$ Therefore, you need to show that for all $n\ge 0$, that the coefficient of $x^n$ in the product $$ (x^0)(x^0+x^1)(x^0+x^2+x^4)\cdots(x^0+x^{k!}+x^{2k!}+\dots+x^{k\cdot k!})\cdots $$ is at least one. Each of those factors in that infinite product is a finite geometric series, which can be simplified, allowing you to evaluate the infinite product...