I am working on this question:
The factorial of a positive integer $n$ is defined by $$n!=1\cdot 2\cdots n.$$ b) Prove that every positive integer $m$ can be written in 'base factorial', i.e. in the form $$m=a_1\cdot 1!+a_2\cdot 2!+\ldots+a_n\cdot n!$$ for some positive integer $n$ and integers $a_1, \ldots, a_n$ which have the property that $0\le a_i \le i$ for each $i\in\{1, \ldots, n\}$.
And trying to use induction with base case $m=1$. Therefore, $n=1$, $a_1=1$.
Inductive hypothesis: Assume that $$m=a_1\cdot 1!+a_2\cdot 2!+\ldots+a_n\cdot n!,$$ we show that $$m+1=a_1\cdot 1!+a_2\cdot 2!+\ldots+a_n\cdot n!.$$ Is this what I am to show?
From there, $$m+1=a_1\cdot 1!+a_2\cdot 2!+\ldots+a_n\cdot n!+1,$$ and I'm not sure how to proceed. I understand this is probably straightforward, but I'm new to this and would appreciate some help. Thanks.
Call a factorial representation "proper" if $0 \le a_i \le i$ for each $i$, and "improper" if all $a_i \ge 0$ and some $a_i \ge i+1$. Clearly every positive integer has some factorial representation (e.g., $m= m\cdot 1!$). You just need to show that there's always a proper one. But note that if you have an improper representation of $m$ in which $a_i \ge i+1$, you can preserve the sum by decreasing the too-large coefficient in the $i$-th place by $i+1$ and increasing the coefficient in the $(i+1)$-th place by $1$. This reduces the "total impropriety" (the sum of $a_i - i$ over the terms where $a_i > i$) by at least $1$... so just keep doing it until you have a proper representation of $m$.