Show that $e^x > \sum_{n=0}^N \frac{x^n}{n!}$

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Knowing that $e^x > 1+x$, I'm supposed to use induction to prove that $ e^x > \sum_{n=0}^N \frac{x^n}{n!}$ when $x > 0$.

The basic step is fairly simple, for $n=1$ the sum equals to $1+x$, and we have proven that.

I then assume it holds for $n = k$, and have: $$ e^x > \sum_{n=0}^k \frac{x^n}{n!} $$

and therefor for $n = k+1$, I have: $$ e^x > \sum_{n=0}^k x^n/n! + \frac{x^{k+1}}{(k+1)!}$$

The problem is I have no idea how I go forward. How do I use it to prove it?