Is this proof of $\sum_{i = 1}^n a_i^k \leq (\sum_{i = 1}^n a_i)^k$ correct? I came across the following proof, and although I believe the result, something seems fishy and I can't put my finger on it.  The base case might not be enough, or we might have to consider various $k$ somewhere...or maybe I'm just paranoid !  So, does that make sense ?
Given a sequence $(a_1, \ldots, a_n)$ of positive integers, we show 
$$\sum_{i = 1}^n a_i^k \leq (\sum_{i = 1}^n a_i)^k$$
for any positive integers $n$ and $k$, by induction on $n$.
If $n = 1$, then obviously $a_1^k \leq a_1^k$.  So assume truth for values smaller than $n$.  Then 
$$
\sum_{i = 1}^n a_i^k = \sum_{i = 1}^{n - 1} a_i^k + a_n^k \\
\leq (\sum_{i = 1}^{n - 1} a_i)^k + a_n^k$$
by induction.  Now, setting $x = \sum_{i = 1}^{n - 1} a_i$ we get 
$$(\sum_{i = 1}^{n - 1} a_i)^k + a_n^k = x^k + a_n^k \leq (x + a_n)^k$$
again by induction, and replacing $x$ by its values lets us conclude that 
$$\sum_{i = 1}^n a_i^k \leq (x + a_n)^k = (\sum_{i = 1}^n a_i)^k$$

Alternately let $x_i = \dfrac{a_i}{\displaystyle \sum_{j=1}^n a_j}$, then $0 < x_i < 1$, and $\displaystyle \sum_{i=1}^n x_i = 1$. Thus we have:
$\displaystyle \sum_{i=1}^n x_i^k \leq \displaystyle \sum_{i=1}^n x_i = 1$ since $0 < x_i^k \leq x_i < 1$ for $\forall k \geq 1$
