Remark that $(a^b-a^c)(a-1)\ge 0$ when $ b \ge c$ Hence the maximum occurs when all variables but one are equal to $1$ The maximum is thus $ra-a+a^{k-r+1}$

jred wrote: Given a natural number $k$ and a real number $a (a>0)$, find the maximal value of $a^{k_1}+a^{k_2}+\cdots +a^{k_r}$, where $k_1+k_2+\cdots +k_r=k$ ($k_i\in \mathbb{N} ,1\le r \le k$). since $a^x$ is convex and $k_{i} \in N$ we have {$k-r+1,1,\cdots ,1$} $\succ$ {$k_{1},\cdots ,k_{n}$} so the claim given above follows by Karamata inequality

The problem isn’t finished since r is not a constant. Note that if $f(r)=(r-1)a+a^{k-r+1}$ $f’(r)=a-lna \cdot{ a^{k-r+1} }$ $f''(r)=(lna)^2 \cdot {a^{k-r+1}} > 0$ So f(r) is a convex function. The maximum is the larger one of $k \cdot a $ and $a^k$.