1. Given $k \in \mathbf{N}$ and a real number $a>0$, let $k_{1}, k_{2}, \cdots, k_{r}$ satisfy the following conditions: $k_{1}+k_{2}+\cdots+k_{r}=k$, $k_{i} \in \mathbf{N}, 1 \leqslant r \leqslant k$. Find the maximum value of $a^{k_{1}}+a^{k_{2}}+\cdots+a^{k}$. 
(CMO - 8 Problem)
1. Clearly, when $0 < a < 1$, for any $s, t \in \mathbf{N}$, we have $a\left(a^{s-1}-1\right)\left(a^{t-1}-1\right) \geqslant 0$, thus $a^{s}+a^{t} \leqslant a+a^{s+t-1}$.
Using the above inequality, we can obtain $a^{k_{1}}+a^{k_{2}}+\cdots+a^{k} \leqslant (r-1) a + a^{k-(r-1)}$,
Consider the equation $a + a^{m} = a^{m+1}$, then we have $a^{m-1} \cdot (a-1) = 1$, which easily solves to $m = \log_{a}\left(\frac{a}{a-1}\right)$, and it can be seen that:
\[
\left\{\begin{array}{l}
a + a^{m} \leqslant a^{m+1}, \text{ when } m \geqslant \log_{a}\left(\frac{a}{a-1}\right), \\
a + a^{m} \geqslant a^{m+1}, \text{ when } m \leqslant \log_{a}\left(\frac{a}{a-1}\right).
\end{array}\right.
\]
Using (2) recursively, we get
\[
(r-1) a + a^{k-(r-1)} \leqslant \left\{\begin{array}{l}
a^{k}, \text{ when } r \leqslant (k+1) - \log_{a}\left(\frac{a}{a-1}\right), \\
k a, \text{ when } r \geqslant (k+1) - \log_{a}\left(\frac{a}{a-1}\right).
\end{array}\right.
\]

From (1) and (3), we get $a^{k_{1}} + a^{k_{2}} + \cdots + a^{k} \leqslant \max \left\{a^{k}, k a\right\}$. (4) Clearly, the two values in the right-hand side of (4) can be achieved, so the maximum value sought is
\[
\max \left\{k a, a^{k}\right\} = \left\{\begin{array}{l}
k a, \text{ when } a \leqslant k^{\frac{1}{k-1}}, k \geqslant 2; \\
a^{k}, \text{ when } k = 1 \text{ or } a > k^{\frac{1}{k-1}}, k \geqslant 2.
\end{array}\right.
\]