The proof of $e^x \leq x + e^{x^2}$

Note the inequality $e^t \ge t + 1$ for all $t \in \mathbb{R}$. In particular $e^{x^2} \ge x^2 + 1$.

If $x \le -1$, then $$ e^{x^2} - e^x + x \ge x^2 + 1 - e^0 + x = x(x+1) \ge 0. $$ If $-1 < x < 1$, then \begin{align*} e^{x^2} - e^x + x &\ge x^2 + x + 1 - \sum_{k \ge 0} \frac{x^k}{k!} \\ &= \frac{x^2}{2!} - \sum_{k \ge 3} \frac{x^k}{k!} \\ &\ge \frac{x^2}{2!} - \sum_{k \ge 3} \frac{x^2}{k!} \\ &= x^2 \left( \frac12 - \left[e - 1 - \frac{1}{1!} - \frac{1}{2!} \right]\right) \\ &= x^2 \left( 3 - e \right) \ge 0. \end{align*}

Finally, if $x \ge 1$, then $$ e^{x^2} - e^x + x > e^{x} - e^x + x > 0. $$