I think induction on $n$ would do the trick here, which means you need to show it holds for $n=2$; note that $U_2\geq 0$ is equivalent to showing $(1+x)^2\geq 1+2x$, which holds for all $x$, whether or not they are greater than $-1$.
Now, assume that $U_n\geq 0$; that is, that $(1+x)^n\geq 1+nx$, and that $x\gt -1$. We want to prove that $U_{n+1}\geq 0$ (i.e., that $(1+x)^{n+1}\geq 1+(n+1)x$).
Take $(1+x)^n\geq 1+nx$. Since $x\gt -1$, then $1+x\gt 0$; multiplying both sides by $1+x$ we get: $\begin{align*} (1+x)^n &\geq 1+nx\\ (1+x)^n(1+x)&\geq (1+nx)(1+x)\\ (1+x)^{n+1}&\geq 1 + nx + x + nx^2\\ (1+x)^{n+1}&\geq 1+ (n+1)x + nx^2 \geq 1+(n+1)x \end{align*}$ with the last inequality since $x^2\geq 0$, so $nx^2\geq 0$. This proves that if $(1+x)^n\geq 1+nx$ and $x\gt -1$, then $(1+x)^{n+1}\geq 1+(n+1)x$. Since $(1+x)^2\geq 1+2x$, then the result holds for all $n$ by induction.
(This can also be done by calculus, noting that $f(x)=(1+x)^n$ lies above its tangent at $x=0$ on the interval $(-1,\infty)$; the tangent at $0$ is precisely $y=1+nx$.)