So this problem appeared in my combinatorics and analysis classes in the past week and I came up with a rather unconventional approach I wanted to ensure is valid
$$(1+x)^n\ge 1+nx\quad \forall \;x>-1,n\in \mathbb N$$
We note that
$$(1+x)^n=\sum_{i=0}^{n}\binom{n}{i}x^{n-i}=\sum_{i=2}^{n}\binom{n}{i}x^{i}+nx+1\ge nx+1$$
Let's use the following notation:
$$s_n(x)=\sum_{i=2}^n\binom{n}{i}x^i$$
We must show that $s_n\ge 0\quad\forall\;x\in(-1,\infty)$
We then note the following:
$$\frac{d^2}{dx^2}\left[\sum_{i=2}^n\binom{n}{i}x^i\right]=s_n\hspace{0.1mm}''(x)=P(n,2)\cdot (x+1)^{n-2}$$
We see this expression has one root at $x=-1$ with multiplicity $n-2$.
Clearly $s_n'' >0\;\forall\;x\ge 0$. Since the second derivative has a single zero, we can then deduce that $s_n''$ is positive on $(-1,0)$.
This implies $s_n$ is convex on $(-1,\infty)$. Since $s_n(0)=0$, this means $s_n$ must be positive on $(-1,\infty)$.
This completes the proof
I don't see anything wrong with your proof. Here's a possible shortcut seeing you're familiar with calculus.
Let $$f(x)=(1+x)^n-nx-1$$ Then $$f'(x)=n(1+x)^{n-1}-n>0\Leftarrow (1+x)^{n-1}>1$$ which is true as $x>-1$. Hence $f(x)$ is monotonically increasing, and since $f(-1)=n-1\ge0$, $$f(x)>0\implies \boxed{(1+x)^n>1+nx}$$ for $n\in\mathbb{N}$ and $x>-1$.