Prove every $m \ge 1\;\exists\;n_m$ so every $n \ge n_m\;\exists\;a_i \in \mathbb{N}$ with $\sum_{i=1}^{n}a_i^{-m}=1$ 
Prove that for every positive integer $m$ there exists a positive
integer $n_m$ such that for every positive integer $n \ge n_m$, there
exist positive integers $a_1, a_2, \ldots, a_n$ such
that$$\frac{1}{a_1^m}+\frac{1}{a_2^m}+\ldots+\frac{1}{a_n^m}=1.$$

''Solution:'':
The "grand" idea of this question is to use Bèzout's identity. So note that $\sum_{i=1}^{k^m}\frac{1}{(ak)^m}=\frac{1}{a^m}$. Then, in fact, if $a=1$, we can write $1$ as the sum of $k^m$ numbers of the form $\frac{1}{k^m}$:
$1=\sum_{i=1}^{k^m}\frac{1}{(k)^m}$, with $n_0=k^m$.
Now, using the last idea, we can write $\frac{1}{k^m}$ as $\sum_{i=1}^{p^m}\frac{1}{(pk)^m}$ (1), adding $p^m -1$ terms to this sum or write $\frac{1}{k^m}$ as $\sum_{i=1}^{q^m}\frac{1}{(qk)^m} $(2), adding $q^m -1$ terms to the sum. Then, the resulting $n$ is equal to: $k^m +(p^m-1)u +(q^m-1)v$, after replacing (1) $u$ times and (2) $v$ times.
We only need to have $\gcd(p^m-1,q^m-1)=1$, because for $\gcd(A,B)=1$ and $a,b$ varying on non-negative integers, then $n= aA+bB$ always has a solution if $n>AB-A-B$ [Chicken McNugget Theorem].
Then we just need to have $\gcd(p^m - 1,q^m-1)=1$. Take $q=l(p^m -1)$ for some positive integer $l$ and it is over, as desired.
Correct?
If correct, would you have an easier way to prove it?
 A: What you've done is basically correct except your proof is not quite complete. The issue is it requires $u + v \le k^m$. However, no matter how large $k$ is, as $n$ gets large enough, this inequality will not always hold. Nonetheless, it's relatively easy & simple to handle this problem.
First, as you noted about the coin problem (with the Chicken McNugget Theorem being a special case, which is applicable here, of where there are $2$ terms), for any choice of integers $p \gt 1$ and $q \gt 1$ so $\gcd(p^m - 1, q^m - 1) = 1$, set
$$A = p^m - 1, \; B = q^m - 1, \; j = AB - A - B \tag{1}\label{eq1A}$$
With any integer $k_0$, have
$$k_i = k_0 + i \; \forall \; i \ge 0 \tag{2}\label{eq2A}$$
Next, set $i = 0$ and choose a $k_0$ large enough so, for each
$$k_i^m + j + 1 \le n \le k_{i+1}^m + j \tag{3}\label{eq3A}$$
there exist non-negative integers $u$ and $v$ where
$$n = k_i^m + (p^m - 1)u + (q^m - 1)v \tag{4}\label{eq4A}$$
and
$$u + v \le k_i^m \tag{5}\label{eq5A}$$
As an exercise for you to do, show for all $i \gt 0$ that, for each $n$ satisfying \eqref{eq3A}, there are non-negative integers $u$ and $v$ so \eqref{eq4A} and \eqref{eq5A} are both true.
This proves, with $n_m = k_0^m + j + 1$, that for all $n \ge n_m$ there exists positive integers $a_i \; \forall \; 1 \le i \le n$ such that
$$\frac{1}{a_1^m} + \frac{1}{a_2^m} + \ldots + \frac{1}{a_n^m} = 1 \tag{6}\label{eq6A}$$
As for there being any easier way to prove \eqref{eq6A}, I don't know of any offhand.
