For every $n$ large enough every zero of $1+z+z^2/2!+\dots z^n/n!$ is such that $|z|>r$ We need to show given $r>0$ there exists $n_0$ such that if $n\ge n_0$ then $1+z+z^2/2!+\dots z^n/n!$ has all $0$'s in $|z|>r$
I was thinking of using Rouche's Theorem  but not able to construct $f(z)$, $g(z)$ with $|f(z)|>|g(z)|$, $|z|>r$ so that $f$, $f+g$ has same no of roots outside $|z|>r$.

No complex analysis is required, only the fact that $f:z\mapsto\mathrm e^z$ has no zero and that the convergence of $f_n:z\mapsto\sum\limits_{k\leqslant n}\frac{z^k}{k!}$ to $f$ is uniform on bounded subsets. 
For every $r$, $f_n$ converges uniformly to $f$ on the closed disk $D(0,r)$. Since $f$ is continuous and has no zero in $D(0,r)$, $|f|\geqslant2\varepsilon$ uniformly on $D(0,r)$, for some $\varepsilon\gt0$. 
And $|f_n-f|\leqslant\varepsilon$ uniformly on $D(0,r)$ for every $n\geqslant N(r,\varepsilon)$. Thus, for every $n\geqslant N(r,\varepsilon)$ and every $|z|\leqslant r$, $|f_n(z)|\geqslant|f(z)|-|f_n(z)-f(z)|\geqslant\varepsilon$, in particular $f_n(z)\ne0$.
