Run the Serre spectral sequence
\begin{equation*}
E^2_{p,q} = H_p\bigl(B;\, H_q(S^n)\bigr) \;\Longrightarrow\; H_{p+q}(E).
\end{equation*}
Since \(H_q(S^n) = \mathbb{Z}\) for \(q = 0, n\) and is zero otherwise, the \(E^2\) page is concentrated in two rows:
\begin{equation*}
E^2_{p,q} \;=\; \begin{cases} H_p(B) & q = 0, n, \\ 0 & \text{otherwise}. \end{cases}
\end{equation*}
The only possibly nonzero differential is
\begin{equation*}
d^{n+1} \colon E^{n+1}_{p,0} = H_p(B) \;\longrightarrow\; E^{n+1}_{p-n-1,\, n} = H_{p-n-1}(B),
\end{equation*}
because \(d^r\) has bidegree \((-r, r-1)\) and only \(r = n+1\) moves from the bottom row to the top row. All earlier \(d^r\) vanish (insufficient vertical reach), so \(E^{r}_{p,q} = E^2_{p,q}\) for \(r \leq n+1\text{,}\) and all later \(d^r\) vanish (one of source/target is in a zero row). Hence
\begin{equation*}
E^\infty_{p,q} \;=\; \begin{cases} \ker\bigl(d^{n+1} \colon H_p(B) \to H_{p-n-1}(B)\bigr) & q = 0, \\ \mathrm{coker}\bigl(d^{n+1} \colon H_{p+n+1}(B) \to H_p(B)\bigr) & q = n, \\ 0 & \text{otherwise}. \end{cases}
\end{equation*}
This rephrases as a four-term exact sequence
\begin{equation}
0 \to E^\infty_{p,0} \to H_p(B) \xrightarrow{\,d^{n+1}\,} H_{p-n-1}(B) \to E^\infty_{p-n-1,n} \to 0.\tag{19.1}
\end{equation}
\begin{equation*}
\{0\} \subseteq F_0 H_i(E) \subseteq \dots \subseteq F_{i-1} H_i(E) \subseteq F_i H_i(E) = H_i(E)
\end{equation*}
with \(E^\infty_{p,q} \cong F_p H_{p+q}(E) / F_{p-1} H_{p+q}(E)\text{.}\) In our case only the rows \(q = 0, n\) are nonzero, so the filtration on \(H_i(E)\) has just two non-trivial steps: \(F_{i-n} H_i(E) = E^\infty_{i-n,n}\) and \(H_i(E) / F_{i-n} H_i(E) = E^\infty_{i,0}\text{.}\) This gives a short exact sequence
\begin{equation}
0 \to E^\infty_{i-n,\,n} \to H_i(E) \to E^\infty_{i,0} \to 0.\tag{19.2}
\end{equation}
Splicing
(19.1) (with
\(p = i\)) and
(19.2) together — i.e. replacing each
\(E^\infty\) in
(19.2) by what
(19.1) tells us it is — yields the desired long exact sequence:
The connecting map
\(d^{n+1}\) on
\(H_*(B)\) is, up to sign, cap product with the Euler class of the sphere bundle.
