Let T_1(x) and T_2(x) be the degree- n Taylor polynomials
Let $T_1(x)$ and $T_2(x)$ be the degree-$n$ Taylor polynomials of $f(x)$ and $\varphi(x)$ respectively, expanded about $x=a$. Prove that the degree-$n$ Taylor polynomial $T(x)$ of the product $y=f(x)\varphi(x)$ is $T(x)=T_1(x)T_2(x)-P(x)$, where $P(x)$ consists of all terms in the product $T_1(x)T_2(x)$ whose degree (in $x-a$) exceeds $n$.
Explanation of Method
Multiply the two truncated series, then simply throw away everything of degree above n.
Step 1: The defining property of a Taylor polynomial
Write $h=x-a$. By definition of the degree-$n$ Taylor polynomial: \[f(x)=T_1(x)+o(h^n), \qquad \varphi(x)=T_2(x)+o(h^n), \tag{1}\] and conversely, a polynomial $Q(x)$ of degree $\le n$ is the degree-$n$ Taylor polynomial of $g$ at $a$ precisely when $g(x)-Q(x)=o(h^n)$.
Step 2: Multiply the two expansions
\[ f(x)\varphi(x) = \bigl[T_1(x)+o(h^n)\bigr]\bigl[T_2(x)+o(h^n)\bigr] = T_1(x)T_2(x)+T_1(x)\,o(h^n)+T_2(x)\,o(h^n)+o(h^{2n}). \tag{2} \]
Since $T_1,T_2$ are bounded near $a$ (they are polynomials), the cross terms are again $o(h^n)$: \[f(x)\varphi(x) = T_1(x)T_2(x) + o(h^n). \tag{3}\]
Step 3: Split the product by degree
The product $T_1(x)T_2(x)$ is a polynomial of degree $\le2n$; split it into its part $S(x)$ of degree $\le n$ and its part $P(x)$ of degree $\gt n$: \[T_1(x)T_2(x) = S(x)+P(x), \qquad S(x)=\!\!\sum_{\substack{j+k\le n\\ j,k\ge0}}\!\! c_jd_k\,h^{j+k}, \qquad P(x)=\!\!\sum_{\substack{j+k>n\\ j,k\ge0}}\!\! c_jd_k\,h^{j+k}, \tag{4}\] where $c_j,d_k$ are the coefficients of $T_1,T_2$ in powers of $h$. Every term of $P(x)$ has degree at least $n+1$, so $P(x)=O(h^{n+1})=o(h^n)$.
Step 4: Identify $S(x)$ as the Taylor polynomial of the product
Substituting (4) into (3): \[f(x)\varphi(x) = S(x)+P(x)+o(h^n) = S(x)+o(h^n). \tag{5}\]
By the characterization in Step 1, $S(x)$ is the degree-$n$ Taylor polynomial of $f\varphi$ at $a$; and since Taylor polynomials of a given degree are unique, $T(x)=S(x)=T_1(x)T_2(x)-P(x)$.
Step 5: Verification
\[a=0,\ n=3,\ f=1+x,\ \varphi=e^x\text{: }T_1T_2=(1+x)(1+x+\tfrac{x^2}2+\tfrac{x^3}6)\text{, discarding degree}\gt 3\text{ terms reproduces the true degree-3 Maclaurin polynomial of }(1+x)e^x\text{ exactly}\ \checkmark\]
Result
Final answer:
\[\boxed{\,T(x) = T_1(x)\,T_2(x) - P(x)\,}\]
where $P(x)$ is the sum of all terms $c_jd_k(x-a)^{j+k}$ in the product with $j+k>n$.