5. Consider all possible sums $c_{1} a_{1}+c_{2} a_{2}+\cdots+c_{n} a_{n}$, where each $c_{i}$ is an integer with $0 \leq c_{i}<m$. There are $m^{n}$ such sums, and if any two of them give the same remainder modulo $m^{n}$, say $\sum c_{i} a_{i} \equiv \sum d_{i} a_{i}\left(\bmod m^{n}\right)$, then $\sum\left(c_{i}-d_{i}\right) a_{i}$ is divisible by $m^{n}$, and since $\left|c_{i}-d_{i}\right|<m$, we are done. We claim that two such sums must exist. Suppose to the contrary that the sums $\sum_{i} c_{i} a_{i}\left(0 \leq c_{i}<m\right)$ give all the different remainders modulo $m^{n}$. Consider the polynomial
$$P(x)=\sum x^{c_{1} a_{1}+\cdots+c_{n} a_{n}},$$
where the sum is taken over all $\left(c_{1}, \ldots, c_{n}\right)$ with $0 \leq c_{i}<m$. If $\xi$ is a primitive $m^{n}$ th root of unity, then by the assumption we have
$$P(\xi)=1+\xi+\cdots+\xi^{m^{n}-1}=0$$
On the other hand, $P(x)$ can be factored as
$$P(x)=\prod_{i=1}^{n}\left(1+x^{a_{i}}+\cdots+x^{(m-1) a_{i}}\right)=\prod_{i=1}^{n} \frac{1-x^{m a_{i}}}{1-x^{a_{i}}},$$
so that none of its factors is zero at $x=\xi$ because $m a_{i}$ is not divisible by $m^{n}$. This is obviously a contradiction. Remark. The example $a_{i}=m^{i-1}$ for $i=1, \ldots, n$ shows that the condition that no $a_{i}$ is a multiple of $m^{n-1}$ cannot be removed.