Here's a simple proof:
$\{\cdot\}$ denotes the fractional part.
$\log_{10}2$ is irrational, so $\{n\log_{10}2\}$ is evenly distributed across $[0,1)$. The leading digit of $2^n$ is $\lfloor10^{\{n\log_{10}2\}}\rfloor$, so it is $x$ iff $\{n\log_{10}2\}$ is in the interval $[\log_{10}x,\log_{10}(x+1))$. This interval has a size of $\log_{10}\frac{x+1}x$, so this is the probability that the first digit of $2^n$ is $x$.
Edit: (Jonathan Allan) - Since this question is on hold I'll put this more expanded version here for the time being.
The basics first, logarithmic and exponentiation identities...
$$x=b^{\log_{b}(x)}\tag{1}\label{1}$$
since logarithm and power functions are inverses.
$$b^{c+d}=b^cb^d\tag{2}\label{2}$$
since multiplication is distributes over addition, i.e:
$$b\times\cdots(c+d\text{ times})\cdots\times b=(b\times\cdots(c\text{ times})\cdots\times b)\times (b\times\cdots(d\text{ times})\cdots\times b)$$
$$\left(b^e\right)^f=b^{ef}\tag{3}\label{3}$$
since multiplication is distributes over multiplication, i.e:
$$(b\times\cdots(e \text{ times})\cdots\times b)\times\cdots(f \text{ times})\cdots\times(b\times\cdots(e \text{ times})\cdots\times b)$$
$$=b\times\cdots(e\times f \text{ times})\cdots\times b$$
Applying a couple of these identities...
By substituting using $(1)$ for $x$:
$$\log_b(x^p)=\log_b((b^{\log_{b}(x)})^p)$$
then applying $(3)$
$$\log_b(x^p)=\log_b({b^{p\log_{b}(x)}})$$
and hence
$$x^p={b^{p\log_{b}(x)}}$$
which means
$$2^n={10^{n\log_{10}(2)}}\tag{A}\label{A}$$
Now an excursion
...into what a written number means in terms of relevance to the puzzle...
A number in a base is an expression of sums of powers
in base $10$ we talk of units, tens, hundreds, thousands and so on (going right to left)
as such writing a number like $743$ is really expressing it in this way:
$743=7\times 100+4\times 10+3\times 1$ or
$743=7\times 10^2+4\times 10^1+3\times 10^0$
We could also move everything into the highest power of $10$:
$743=7.43\times 10^2$
Such that the number on the left of the decimal point is the leading digit and is therefore in $[1,9]$
Now we can note that $\log_b{1}=0$ and $\log_b{b}=1$ (or equivalently $b^0=1$ and $b^1=b$) so
$743=10^r10^2$
with $r\in[0,1)$, that is $r$ is between $0$ (inclusive) and $1$ (exclusive)
- for this example of $743$, $r=0.87098881376\dots$
hence
$7.43=10^r$
and
$7=\lfloor10^r\rfloor$ (the integer part of) - which is the leading digit of the number $743$.
Similarly
for the problem at hand where $n$ is a positive integer:
$$2^n=10^r10^s$$
with $s\in\Bbb{N}$ (a positive integer or zero) and $r\in[0,1)$
But by $(2)$ in reverse that means
$$2^n=10^{r+s}$$
So $r$ is the fractional part of $r+s$
and from $(A)$:
$$r+s=n\log_{10}(2)$$
Therefore we have that the leading digit in base $10$ of $2^n$ is
$$\lfloor10^{\{n\log_{10}(2)\}}\rfloor\tag{B}\label{B}$$
where $\{\cdot\}$ represents taking the fractional part.
Now
if $\log_{10}(2)$ were rational it would be expressible as $\frac tu$ with $t,u\in \Bbb{N^*}$ ($t$ and $u$ positive integers)
Furthermore since $1<2<10$ we know that (as we knew for $r$)
$$\log_{10}(2)\in[0,1)$$
so $u>t$
But\begin{align}\log_{10}(2)&=\frac tu\\\rightarrow u\log_{10}(2)&=t\\\rightarrow\log_{10}(2^u)&=t\\\rightarrow 2^u&=10^t\\\rightarrow2^u&=2^t5^t\\\rightarrow2^{u-t}&=5^t\end{align}
Which would make an even number equal an odd number, so $\log_{10}(2)$ cannot be rational
Hence $\log_{10}(2)$ is irrational.
Although
...I'm not sure of a simple proof of the equidistribution theorem but for an elementary proof see this post on MathOverflow (however, it does seem intuitive due to the ever changing landscape of decimal digits of an irrational number)
This tells us that $\{n\log_{10}(2)\}$ will be uniformly distributed (by choice of $n$) across the interval it may occupy, $[0,1)$, as $n$ gets large because $\log_{10}(2)$ is irrational.
Finally
... by $(B)$ the chance of a leading digit of $2^n$ being $d$ is the chance of $\lfloor10^{\{n\log_{10}(2)\}}\rfloor$ being $d$
...and if
$$d=\lfloor10^{\{n\log_{10}(2)\}}\rfloor$$
then
$$d\leq 10^{\{n\log_{10}(2)\}} < d + 1$$
so (since logarithm and power functions are inverses)
$$\log_{10}(d)\leq \{n\log_{10}(2)\} < log_{10}(d + 1)$$
Which neatly splits up the range $[0,1)$ (since $\log_{10}(1)=0$ and $\log_{10}(10)=1$)
Leaving the chance of a leading digit of $2^n$ being $d$ being the fraction of the range $[0,1)$ occupied by the range $[\log_{10}(d),\log_{10}(d+1))$
Thus the probability of a leading digit being $d$ is
$$P(d)=(\log_{10}(d+1)-\log_{10}(d))/(1-0)$$
which by the inverse of $(2)$ means
$$P(d)=\log_{10}(\frac{d+1}d)\space\space\space\space\Box$$
To answer the gambling question
Your friend's best strategy is to pick $d$ such that his expected value$$\$(d+2)P(d)-1$$ is maximised, which is the same as maximising $$(d+2)P(d)=(d+2)\log_{10}(\frac{d+1}d)=\log_{10}({\frac{d+1}d}^{d+2})$$which is the same as maximising$$\frac{d+1}{d}^{d+2}={(1+\frac 1d)^{d+2}}=1+\frac 1d^{d+2}=1+d^{-d-2}$$which decreases as $d$ increases, so he should pick $d=1$, whereupon his expected value is$$\$(1+2)P(1)-1=\$3\log_{10}(2)-1=\$\log_{10}(2^3)-1=\$\log_{10}(8)-1$$Since $$x<b\rightarrow\log_b{x}<1$$ his expectation is negative, and since the game is zero sum our expectation of $$\text{our EV}=\$1-\log_{10}(8)$$is positive
We don't have a calculator in our bathroom, but that is about $9.69\text{c}$ per game played.
(Note that if he goes for the longest odds of $10:1$ by choosing $d=9$ every time our expected value is around $49.67\text{c}$ per game played.)
