Proving the odd Bernoulli numbers are zero suppose we define the Bernoulli numbers $b_n, n = 1, 2, 3, \ldots$ by the Faulhaber's fomula 
$$\begin{eqnarray*}1 + 2^k + 3^k + \ldots n^k &=& \frac{1}{k+1}[n^{k+1} + b_1 c(k+1,2) n^k + b_2 c(k+1,3)n^{k-1} + \ldots ]\\&=& \frac{(n+b)^{k+1}-b^{k+1}}{k+1}\end{eqnarray*}$$
with the proviso we replace $b^k$ by $b_k$ in the binomial expansion and $c(k,l)$ is $k$ choose $l.$ 
my question is, how do you show all the odd Bernoulli numbers except $b_1$ is zero without invoking heavy analytical tools so that the reason can be explained to
a student who has only had one or two semesters of calculus in high school.

thanks for all the responses. i will collect the results. this is too large to fit into a comment to quanta's response.
we will use $\sigma_k = 1^k + 2^k + \ldots n^k$
proving the odd bernoulli constants $b_k$ are zero for $= k = 3, 5, \ldots.$
let us fix the notation and collect some elementary results.
$\sigma_k = 1^k + 2^k + \ldots n^k, \sigma_1 = \frac{n(n+1)}{2}, \sigma_3 = \sigma_1^2.$
the idea of the proof is
to show that $n^2$ divides $\sigma_k$ for all $k \ge 3$ by induction. clearly true for $k = 3.$
we will show that $\sigma_k$ is a linear combination of $\sigma_{k-2}, \ldots, \sigma_3$ and
$\sigma_1^k$ for all $k \ge 3.$ that concludes the proof.
now i will establish the needed linear dependence of $\sigma_k$ which is 
$$\sigma_1^k =    \frac{1}{2^{k-1}} 
  \cases{[ {k \choose 1}\sigma_{2k-1} + {k \choose 3}\sigma_{2k-3} +\ldots +\sigma_k ]
   & for $k$ odd, \cr
         [ {k \choose 1}\sigma_{2k-1} + \ldots + {k \choose 3}\sigma_{2k-3} ]
   & for $k$ even.}
$$
i will show all the intermediate steps for $k = 5.$
it is easy to see the steps for the $k = 5$ carry over to general case.
\
$\begin{array}
\sigma_1^5 = (1+2\ldots n)^5 = \frac{n^5(n+1)^5}{2^5} \\
~~~~  = \frac{1}{2^5}\{[n^5(n+1)^5 -(n-1)^5n^5] + [(n-1)^5n^5 -(n-2)^5(n-1)^5]
  +\ldots +[2^53^5 - 1^52^5] + 1^5 2^5 \}\\
~~~~  = \frac{1}{2^5}\{ n^5[(n+1)^5 -(n-1)^5] + ((n-1)^5[n^5 -(n-2)^5]
+\ldots +2^5[3^5 - 1^5] + 1^5 2^5 \}\\
~~~~ =  \frac{1}{2^5}\{ 2n^5[{5 \choose 1}n^4 + {5 \choose 3}n^2 + 1] +
                        2(n-1)^5[{5 \choose 1}(n-1)^4 + {5 \choose 3}(n-3)^2 + 1]+\\
~~~~~~        \ldots +  2^5[{5 \choose 1}2^4 + {5 \choose 3}2^2 + 1] + 
                        1^5[{5 \choose 1}1^4 + {5 \choose 3}1^2 + 1]  
\}\\
~~~~~~ = \frac{1}{2^4}\{ {5 \choose 1}[n^9 + (n-1)^9 + \ldots + 2^9 + 1^9] +
                         {5 \choose 3}[n^7 + (n-1)^7 + \ldots + 2^7 + 1^7] +
                                      [n^5 + (n-1)^5 + \ldots + 2^5 + 1^5]
\}\\
~~~~~ = \frac{1}{2^4}[ {5 \choose 1}\sigma_9 +
                         {5 \choose 3}\sigma_7 +
                                      \sigma_5 ].
\end{array}
$ 
