= Solution
The <splitting principle for complex vector bundles> says that for every complex vector bundle $E\to X$ there is a map $p:F(E)\to X$ such that $p^*$ is injective on cohomology and
$$
p^*E=L_1\oplus\cdots\oplus L_r
$$
splits into complex line bundles. Write $x_i=c_1(L_i)$ for the formal <Chern roots>.
Define the <Chern character> after this injective pullback by
$$
\operatorname{ch}(E)=\sum_{i=1}^r e^{x_i}.
$$
Each homogeneous component is a symmetric polynomial in the $x_i$ with rational coefficients, hence a polynomial in the elementary symmetric functions $c_j(E)$. It therefore descends uniquely to $H^{\mathrm{ev}}(X;\mathbb Q)$ and depends only on $E$. Set
$$
\operatorname{ch}(E-F)=\operatorname{ch}(E)-\operatorname{ch}(F)
$$
on the <Grothendieck group> $K^0(X)$; additivity under direct sums makes this well defined.
If $E$ has roots $x_i$ and $F$ has roots $y_j$, then $E\otimes F$ has roots $x_i+y_j$. Consequently
$$
\begin{aligned}
\operatorname{ch}(E\oplus F)
&=\sum_i e^{x_i}+\sum_j e^{y_j}
=\operatorname{ch}(E)+\operatorname{ch}(F),\\
\operatorname{ch}(E\otimes F)
&=\sum_{i,j}e^{x_i+y_j}
=\left(\sum_i e^{x_i}\right)
\left(\sum_j e^{y_j}\right)
=\operatorname{ch}(E)\operatorname{ch}(F).
\end{aligned}
$$
It also sends the trivial line to $1$, so it is a unital ring homomorphism.
For $S^{2n}$, a generator of $\widetilde K^0(S^{2n})$ is the $n$-fold exterior product of the degree-two <Bott element>. The Chern character respects exterior products, and the degree-two Bott element has Chern character equal, up to sign, to the integral generator of $\widetilde H^2(S^2;\mathbb Z)$. Its $n$-fold product maps to the integral top-dimensional generator. Hence the <Chern character on an even-dimensional sphere is integral>.
Let the formal Chern roots of $E\to S^{2n}$ be $x_1,\ldots,x_r$ and write $p_n=\sum_i x_i^n$. Since
$$
H^{2j}(S^{2n};\mathbb Z)=0
\qquad(0<j<n),
$$
all lower Chern classes $c_1(E),\ldots,c_{n-1}(E)$ vanish. The <Newton identities> then reduce to
$$
p_n=(-1)^{n+1}n\,c_n(E).
$$
The degree-$2n$ term of the Chern character is therefore
$$
\operatorname{ch}_n(E)
=\frac{p_n}{n!}
=(-1)^{n+1}\frac{c_n(E)}{(n-1)!}.
$$
Its evaluation on the <fundamental class> is an integer by integrality of the reduced Chern character. Thus
$$
\left\langle c_n(E),[S^{2n}]\right\rangle
$$
is divisible by $(n-1)!$, proving the <Divisibility of the top Chern number on an even-dimensional sphere>.
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