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.
 A: Define $\sigma_k(n) = \sum_{i=1}^n i^k$ then since $\sigma_1(n) = \frac{n(n+1)}{2}$
First to show that the coefficient of $n$ in $\sigma_{2k+1}$ is $b^{2k+1}$: It's immediate from applying the binomial theorem to $\sigma_{2k+1}(n) = \frac{(n+b)^{k+1}-b^{k+1}}{k+1}$. Then we can show this coefficient is always zero by showing these polynomials are divisible by something with $n^2$ as its lowest power.
This argument is due to Pascal (I could not possibly come up with it!) taken from here,
$$\begin{array} . \sigma_1(n)^k &=& \sum_{m=1}^n \left[ \left(\frac{m(m+1)}{2}\right)^k - \left(\frac{(m-1)m}{2}\right)^k \right] \\
&=& \sum_{m=1}^n \sum_{r=0}^k \binom{k}{r} \left(\frac{m}{2}\right)^k m^r (1 - (-1)^{k-r}) \\
&=& \frac{1}{2^k} \sum_{r=0}^k \binom{k}{r} \sigma_{k+r}(n) (1 - (-1)^{k-r})
\end{array}$$
Now consider $k$ odd, many of the terms cancel, we are left with the fact that $\sigma_1^k$ is a linear combination of various $\sigma_{i}$ for $i$ odd.
The relation $\sigma_3 = \sigma_1^2$ is classical and can be proved in a variety of simple ways. Induction on (odd) $k$ shows that every other odd one ($\sigma_5$, $\sigma_7$ and so on) is also divisible by $\sigma_1^2$. In fact you can try out these induction arguments concretely for given $k$ to express these summations in terms of triangular numbers. That is called producing the Faulhauber polynomials.
This proves that there is no $n$ terms in these odd ($\ge 3$) sums of powers, but since the coefficient of $n$ in $\sigma_{2k+1}$ is $b_{2k+1}$ this proves that they are all zero.
A: Why not take the opportunity to expose them to the glory of generating functions?
Define the Bernouli numbers to be the coefficients in
$$\frac{t}{1-e^{-t}} = \sum B_m \frac{t^m}{m!}.$$
Prove that these give the sum of the $k$-th powers by equating coefficients of $x^k$ in
$$\sum_{j=1}^n \frac{j^k x^k}{k!} = \sum_{j=1}^n e^{jx} = \frac{e^{(n+1)x} - e^x}{e^x-1}$$ 
$$= \left( \frac{e^{nx} -1}{x} \right) \left( \frac{x}{1-e^{-x}} \right) = \sum_{\ell=0}^{\infty} \frac{n^{\ell+1} x^{\ell}}{(\ell+1)!}  \cdot \sum B_m \frac{t^m}{m!}.$$
Then prove that the odd Bernoulli numbers vanish by noting that
$$\sum B_m \frac{t^m}{m!} - \sum B_m \frac{(-t)^m}{m!} = \frac{t}{1-e^{-t}} - \frac{-t}{1-e^t} = \frac{t (e^t-1)}{e^t-1} = t.$$
A: 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}
$ 
