= Solution
For <càdlàg> <semimartingales>, the <semimartingale integration by parts> formula is
$$
\boxed{X_tY_t=X_0Y_0+\int_0^tX_{s-}\,dY_s+\int_0^tY_{s-}\,dX_s+[X,Y]_t.}
$$
For continuous processes the left limits can be replaced by the values. The bracket includes the jump products in the general formula.
For a continuous $\mathbb R^d$-valued <semimartingale> $X$ and $f\in C^2(\mathbb R^d)$, the <Itô formula> is
$$
\boxed{f(X_t)=f(X_0)+\sum_{i=1}^d\int_0^t\partial_if(X_s)\,dX_s^i+\frac12\sum_{i,j=1}^d\int_0^t\partial_{ij}f(X_s)\,d[X^i,X^j]_s.}
$$
The formula is localized when the derivatives are unbounded; continuity of $X$ makes them bounded after stopping on compact sets.
To prove the one-dimensional formula for polynomials without assuming that formula, use induction. It is immediate for constants and $X$. Suppose
$$
d(X^k)=kX^{k-1}\,dX+\frac{k(k-1)}2X^{k-2}\,d[X].
$$
Write $X=X_0+M+A$ with $M$ a <continuous local martingale> and $A$ <finite variation>. The induction hypothesis identifies the <martingale> part of $X^k$ as $kX^{k-1}\cdot M$. The stochastic-integral covariation identity, obtained by polarization from the quadratic-variation identity, gives
$$
d[X^k,X]=kX^{k-1}\,d[X].
$$
Apply <semimartingale integration by parts> to $X^kX$. Its first-order terms combine to $(k+1)X^k\,dX$, and its bracket terms combine to
$$
\left(\frac{k(k-1)}2+k\right)X^{k-1}\,d[X]=\frac{k(k+1)}2X^{k-1}\,d[X].
$$
This proves the induction step. Taking linear combinations proves the <polynomial Itô formula from integration by parts> for every polynomial $f$.
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