At A-level I'd have regarded this as "why I like maths" because it worked, but now I look at it and think "It doesn't look like it should work" and question that.
I am showing $$\sum^m_{k=1}\text{ }^nC_k a_{k,m}=n^m$$ so that I may show sums of integers raised to powers have nice expressions.
I have concluded $$a_{i,j}=\sum_{k=1}^i(-1)^{(i+k)}\text{ }^iC_kk^j$$
This is correct and truly a lovely bit of maths, but what is funny is that when $i>j$ it is zero. This makes perfect sense based on how I got to it, I do not doubt it is zero it's just looking at it, I can find no reason why it ought to be zero. (If I gave you this, how would you show it is zero, going back the way I came is not.... intuitive, I couldn't make the leap that it was zero, without knowing how I got to it)
Why it is zero (you don't have to read this bit)
You can see it ought to be zero without touching matrices by looking at the first equation, notice that $^nC_k$ will be a polynomial of order k, the coefficient are such that if you expanded the sum with these polynomials, it would tidy to $n^m$. You would not want an order $>m$ in the mix! (it is because $a_{k,m}=0$ for $k>m$ that the first summation stops at m, rather than being a full matrix multiplication). The matrices involved are $C_n$ an $nxn$ matrix with $(C_n)_{i,j}=^iC_j$ multiplied by $A_n$ the $nxn$ matrix of my unknown magic numbers (the $a_{i,m}$s), and that equals what I call the "power matrix", which is the $nxn$ matrix $P_n$ where $(P_n)_{i,j}=i^j$, thus $C_nA_n=P_n$ giving me $A_n=C^{-1}_nP_n$
The inverse of $C^{-1}_n$ is remarkably nice. $(C^{-1}_n)_{i,j}=(-1)^{i+j}(C_n)_{i,j}$ which is how I found this.
The reason that your sum is zero when $i>j$ is that your sum represents the number of surjective mappings from a set with $j$ elements to a set with $i$ elements, and of course there are no such maps when $i>j$.
But why does your sum represent the number of surjective maps? To prove it, use the principle of inclusion exclusion. Suppose $X$ has $j$ elements and $Y$ has $i$ elements. Let $U=Y^X$ be the set of all functions from $X$ to $Y$ and for any $y\in Y$ let $F_y$ denote the set of functions in $U$ which do not contain $y$ in their image. The number of surjective functions is exactly the number of functions not in $\bigcup_{y\in Y} F_y$, and this is $$\sum_{K\subseteq Y} (-1)^{|K|}\Big|\bigcap_{y\in K} F_y\Big| = \sum_{k=0}^i(-1)^k\cdot T(k),$$ where $T(k)$ is the number of functions in $U$ whose image contains exactly $i-k$ elements. A basic counting argument shows that $T(k)=\binom{i}{k}(i-k)^j$. It follows that the number of surjective functions is $$ \sum_{k=0}^i(-1)^k\binom{i}{k}(i-k)^j. $$ Replacing $k$ by $i-k$, this is $$ \sum_{k=0}^i(-1)^{i-k}\binom{i}{i-k}(i-(i-k))^j = \sum_{k=1}^i(-1)^{i+k}\binom{i}{k}k^j. $$