Proof of $1+x\leq e^x$ for all x?

The strict inequality should be $\le$, as pointed out in the comments.

One way is to look at the derivative of $f(x) = e^x - x - 1$, which is $f'(x) = e^x - 1$, and note that it is zero only at $x=0$. The second derivative is $f''(x)=e^x > 0$ for all $x$, so $x=0$ is a global minimizer. Finally, note $f(0) = 0$, which yields $e^x-x-1 = f(x) \ge 0$ for all $x$.

While other answers have tried to give different proofs of the inequality, I will deal with the more important and slightly difficult second part where it asks for minimum value of $a$ such that $a^{x} \geq 1+x$.

Well it turns out that there is one and only one value of $a$ such that $a^{x} \geq 1+x$ and that value is $a=e$. We have the following theorem:

Theorem: Let $f:\mathbb{R} \to\mathbb{R} $ be a function such that $f(x) \geq 1+x$ for all $x\in \mathbb {R} $ and further $f(x+y) =f(x) f(y) $ for all $x, y\in\mathbb {R} $. Then $f(x) =\exp(x)$.

Proof: First some obvious observations. From the inequality $f(x) \geq 1+x$ we can see that $f(x) >0$ for all $x\geq 0$. And putting $x=y=0$ in functional equation we get $f(0)=f(0)f(0)$. Since $f(0)>0$ it follows that $f(0)=1$. Next $f(x) f(-x) =f(0)=1$ and hence if $x>0$ then $f(-x) =1/f(x)>0$. Thus we have proved that $f$ takes only positive values.

Now consider $0<x<1$ and then we have $$\frac{f(x) - 1}{x}\geq 1\tag{1}$$ Next $f(-x) \geq 1-x$ or $1/f(x)\geq 1-x$ or $$f(x) \leq \frac{1}{1-x}$$ or $$\frac{f(x) - 1}{x}\leq \frac{1}{x}\left(\frac{1}{1-x}-1\right)=\frac{1}{1-x}\tag{2}$$ Combining $(1),(2)$ we get $$1\leq \frac{f(x) - 1}{x}\leq\frac{1}{1-x}$$ Letting $x\to 0^{+}$ we get via Squeeze Theorem $$\lim_{x\to 0^{+}}\frac{f(x)-1}{x}=1\tag{3}$$ And because of the above limit we see that $f(x) \to 1$ as $x\to 0^{+}$. Next we have $$\lim_{x\to 0^{-}}\frac{f(x)-1}{x}=\lim_{t\to 0^{+}}\frac{1-f(-t)}{t}=\lim_{t\to 0^{+}}\frac{f(t)-1}{t}\cdot\frac{1}{f(t)}=1$$ Hence we finally have $$f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x}=\lim_{x\to 0}\frac{f(x)-1}{x}=1\tag{4}$$ Using this and the functional equation $f(x+y) =f(x) f(y) $ we can easily prove that $f'(x) =f(x) $ for all $x$. With $f(0)=1$ this uniquely characterizes the exponential function.

For $x \lt -1\,$, obviously $1+x \lt 0 \lt e^x\,$. For $x \ge -1\,$, by Bernoulli's inequality: $$\left(1+\frac{x}{n}\right)^n \ge 1 + n \cdot \frac{x}{n} = 1+x \quad\quad\text{for} \;\;n \ge 1$$

Passing to the limit for $\,n \to \infty\,$ gives $\,e^x \ge 1+x\,$.