[proofplan] We prove non-emptiness of the set of admissible numbers of summands and then use the [well-ordering principle](/theorems/721) to obtain the least one. The input is the asymptotic Waring theorem supplied by Hua's bound: for each fixed $k$, all sufficiently large integers are sums of a bounded number of positive $k$-th powers. The remaining integers form a finite exceptional range, and each exceptional integer is represented by using copies of $1^k$ and padding with zeros. [/proofplan] [step:Use Hua's bound to represent all sufficiently large integers] Fix $k\in\mathbb N$ with $k\ge 2$, and define \begin{align*} H_k:=2^k+1. \end{align*} Let $G(k)$ denote the asymptotic Waring number, namely the least integer $r\in\mathbb N$ such that every sufficiently large positive integer is a sum of exactly $r$ positive $k$-th powers. The theorem [citetheorem:9077] applies because $k\ge 2$, and it gives \begin{align*} G(k)\le H_k. \end{align*} Set \begin{align*} R_k:=G(k). \end{align*} By the definition of $G(k)$, there exists an integer $N_k\in\mathbb N$ such that every integer $n\in\mathbb N$ with $n\ge N_k$ admits a representation \begin{align*} n=y_1^k+\cdots+y_{R_k}^k \end{align*} with $y_1,\dots,y_{R_k}\in\mathbb N$. The inequality $R_k\le H_k$ will allow these representations to be padded by zeros when we pass from positive powers to non-negative powers. [guided] Fix $k\in\mathbb N$ with $k\ge 2$. We introduce the explicit integer \begin{align*} H_k:=2^k+1. \end{align*} Let $G(k)$ denote the asymptotic Waring number, with the following convention: $G(k)$ is the least integer $r\in\mathbb N$ such that every sufficiently large positive integer is a sum of exactly $r$ positive $k$-th powers. The theorem [citetheorem:9077] has the hypothesis $k\ge 2$, which is exactly our present hypothesis, and it states that Hua's method gives \begin{align*} G(k)\le 2^k+1. \end{align*} Thus, with our notation, \begin{align*} G(k)\le H_k. \end{align*} Set \begin{align*} R_k:=G(k). \end{align*} The definition of $G(k)$ does not say that every sufficiently large integer is a sum of exactly $H_k$ positive powers. It says that every sufficiently large integer is a sum of exactly $R_k$ positive powers, where $R_k\le H_k$. Therefore there exists a threshold integer $N_k\in\mathbb N$ such that every $n\in\mathbb N$ with $n\ge N_k$ can be written as \begin{align*} n=y_1^k+\cdots+y_{R_k}^k \end{align*} for some $y_1,\dots,y_{R_k}\in\mathbb N$. The reason this is enough for Waring's theorem is that the final theorem allows non-negative $k$-th powers. Since $0^k=0$, the $R_k$ positive summands can later be padded with zeros until the total number of summands is the fixed number $s_k$. [/guided] [/step] [step:Absorb the finite exceptional range by increasing the number of allowed summands] Define \begin{align*} E_k:=\max\{N_k-1,0\} \end{align*} and \begin{align*} s_k:=H_k+E_k. \end{align*} We prove that this $s_k$ is admissible. Let $n\in\mathbb N$. If $n\ge N_k$, choose $y_1,\dots,y_{R_k}\in\mathbb N$ such that \begin{align*} n=y_1^k+\cdots+y_{R_k}^k. \end{align*} Set $x_i:=y_i$ for $1\le i\le R_k$ and set $x_i:=0$ for $R_k<i\le s_k$. This definition is valid because $R_k\le H_k\le s_k$. Since $0\in\mathbb N_0$, this gives $x_1,\dots,x_{s_k}\in\mathbb N_0$ and \begin{align*} n=x_1^k+\cdots+x_{s_k}^k. \end{align*} If $1\le n<N_k$, then $n\le E_k$. Set $x_i:=1$ for $1\le i\le n$ and set $x_i:=0$ for $n<i\le s_k$. Since $1^k=1$ and $0^k=0$, we obtain \begin{align*} x_1^k+\cdots+x_{s_k}^k=n. \end{align*} Thus every positive integer is a sum of exactly $s_k$ non-negative $k$-th powers. [/step] [step:Take the least admissible number of summands] Define the set of admissible summand counts by \begin{align*} \mathcal A_k:=\{s\in\mathbb N:\text{ every }n\in\mathbb N\text{ is a sum of }s\text{ non-negative }k\text{-th powers}\}. \end{align*} The previous step proves $s_k\in\mathcal A_k$, so $\mathcal A_k$ is a nonempty subset of $\mathbb N$. By the well-ordering principle, $\mathcal A_k$ has a least element. This least element is precisely the Waring number $g(k)$, so $g(k)$ exists. [/step]