Proof of $1+x\leq e^x$ for all x? Does anyone provide proof  of $1+x\leq e^x$ for all $x$?
What is the minimum $a(>0)$  such that $1+x\leq a^x$ for all x?

By using induction I have showded that $$1+n ≤ e^n$$
for $n \in \mathbb{N}$
Induction start $$P(1):1+1≤ e^1$$ 
$$2 ≤ e $$ 
Induction Step
$$P(n):1+n≤ e^n$$ Adding plus 1 to both sides
$$n+2 ≤ e^n+1$$ 
We know now that:$$e^n+1≤ e^{n+1}$$ Since we know that multiplying by a number will yield a higher result compared to adding the same number under conditions: Both numbers have to be positive and cannot be zero.
$$P(n+1):n+2≤ e^{n+1}$$ 
$$P(n+1):1+n+1≤ e^{n+1}$$ 
Concluding, we know that $$f(x) = e^{x}$$ will also be greater than $$g(x) = 1+x$$ but only for $x \in \mathbb{N}$
Now in the second part, I will prove that there is only 1 intersection at x=0.We define a function $$f(x) = e^{x},g(x)=1+x,f-g=e^{x}-(1-x)$$ f-g is the vertical distance function. And now we show that at one point the distance will be zero (Our only intersection point).
Solve (f-g)'= 0
$$(f-g)'= e^{x}-1$$
$$(e^{x}=1$$
$$x=ln(1)$$
$$x=0$$
Now evaluate the function (f-g) for the value 0. 
$$(f-g)(0)=0$$
In the first part using induction I have showed that function f is greater than g for $x \in \mathbb{N}$.
In the second part I have showed that there is only 1 intersection point.
Concluding, we can now say that f is greater than g for $x \in \mathbb{R}$.
One part of the proof is still open: the negative values of x. Hint: Transformation of functions;f(x-k),g(x-k). All arguments still valid.
