<s>[INST] $2^x\leq x+1$ for $x\in [0,1]$ I tried using mean value theorem but couldn't show $2^{2^c} < e$ for $c ∈ (0,x)$. Writing taylor expansion of $2^x$ also don't work because we need to show $2^x$ is smaller than something, not greater
 [/INST] Since other people have explained the typical way to do this, I'll mention that there's a very tricky way to show $2^x \leq 1 + x$ for $0 \leq x \leq 1$ with an infinite series expansion. But not a Taylor series - an infinite series of binomial coefficients.
If we define the binomial coefficients $\binom{x}{k}$ for non-integer arguments $x$ as $$\binom{x}{0} := 1 \text{ and } \binom{x}{k} := \frac{x(x-1)...(x-k+1)}{k!} \text{ for } k \geq 1,$$
then $$2^x = \sum_{k = 0}^\infty \binom{x}{k}$$ is an absolutely convergent series for $x > 0$, and a conditionally convergent series for $0 > x > -1$†, as we have the asymptotic formula $$\binom{x}{k} \approx \frac{(-1)^k}{\Gamma(-x)k^{x+1}}$$ as $k \rightarrow \infty$. Writing out the first few terms explicitly:
$$2^x = \sum_{k = 0}^\infty \binom{x}{k} = 1 + x + \binom{x}{2} + \binom{x}{3} + ...$$
We are now ready to get our inequality $2^x \leq 1 + x$ for $0 \leq x \leq 1$. In fact, we can prove something stronger: $2^x < 1 + x$ for $0 < x < 1$ (i.e. the inequality is strict everywhere except the endpoints).
For $0 < x < 1$ and $k \geq 1$, the terms $\binom{x}{k}$ are nonzero, alternating in sign, and their absolute values are monotone decreasing towards zero, since $$\binom{x}{k+1} = \frac{x-k}{k+1} \binom{x}{k} = (-1)\frac{k-x}{k+1} \binom{x}{k},$$
and $0 < \frac{k-x}{k+1} < 1$ when $k \geq 1$ and $0 < x < 1$.
Therefore, in the range $0 < x < 1$, $$1 + x + \binom{x}{2} + \binom{x}{3} + ...$$ satisfies the conditions of the alternating series test from the second term onwards. Because in an alternating series, the odd partial sums are upper bounds and the even partial sums are lower bounds when the first term (in this case, $1+x$) is positive, we get a whole family of inequalities for $2^x$, among them the one in the original post:
\begin{align*}
2^x & \leq 1 + x \\
2^x & \geq 1 + x + \binom{x}{2} \\
2^x & \leq 1 + x + \binom{x}{2}+\binom{x}{3} \\
2^x & \geq 1 + x + \binom{x}{2}+\binom{x}{3}+ \binom{x}{4} \\
 & ... \\
2^x & \leq 1 + x + \binom{x}{2}+...+\binom{x}{2n-1} \\
2^x & \geq 1 + x + \binom{x}{2}+...+\binom{x}{2n} \\
 &... \
\end{align*}
for all $0 \leq x \leq 1$ and all $n \geq 1$, with equality only if $x = 0$ or $x = 1$.

† When $x$ equals $0$, it is a finite series: $\binom{0}{0} = 1$ and $\binom{0}{k} = 0$ for all other $k$, so $2^0 = \sum_{k = 0}^\infty \binom{0}{k} = 1 + 0 + 0 + ... = 1.$ This is true more generally: for $x$ any nonnegative integer, the series will be finite, with all but its first $x+1$ terms equal to $0$.
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