Prove the roots of these exponential functions are integers? The zeros of $3^x-2^x$ is 0.
The zeroes of $4^x-2\cdot3^x+2^x$ are 0 and 1.
The zeroes of $5^x-3\cdot4^x+3\cdot3^x-2^x$ are 0, 1, and 2.
The zeroes of $6^x-4\cdot5^x+6\cdot4^x-4\cdot3^x+2^x$ are 0, 1, 2, and 3.
The zeroes of $7^x - 5\cdot6^x+10\cdot5^x-10\cdot4^x+5\cdot3^x-2^x$ are 0, 1, 2, 3, and 4.
The zeroes of $8^x-6\cdot7^x+15\cdot6^x-20\cdot5^x+15\cdot4^x-6\cdot3^x+2^x$ are 0, 1, 2, 3, 4, and 5.
And so on.... Basically the pattern here is the base decreases by 1, and the coefficients are of the Pascal's triangle.
So, with infinite terms, this function produces  roots that are the integers. Unlike the trigonometric functions (such as $sin(\pi\cdot x)$ which has roots that are integers, including negative numbers) that have a constrained range between -1 and 1, these series of functions amplitude increases back and forth between the roots until the highest power overpowers the other terms, as can be seen in the image below:



 A: Let $f(t) $ be any function defined over $[0,\infty)$. Its
forward difference is a function defined over same domain by the equation:
$$\Delta f(t) = f(t+1) - f(t)$$
Notice when $f(t)$ is a polynomial of degree $m$, $\Delta f(t)$ is a polynomial of degree $m-1$ (or zero if $f(t)$ is a constant polynomial).
Apply forward difference $n$ times, we find
$$\Delta^nf(t) = \sum_{k=0}^n(-1)^{n-k}\binom{n}{k}f(t+k)$$
vanishes whenever $f(t)$ is a polynomial with $\deg f(t) < n$.
Substitute $f(t)$ by $(t+2)^x$, we obtain:
$$\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}(k+2)^x = 0
\quad\text{ at }\quad x = 0,\ldots, n-1.\tag{*1}$$
This means your equation has at least $n$ solutions in $x$.
The sort of function on LHS of $(*1)$ is called Dirichlet polynomial.
In 1883, Laguerre has proved a theorem${}^{\color{blue}{[1]}}$:

Generalized Descartes' rule of signs
Given a Dirichlet polynomial $P(x) = \sum\limits_{j=0}^n a_j b_j^x$ where
$a_j, b_j \in \mathbb{R}$ satisfies:
$$ a_0, a_1, \ldots, a_n \ne 0,\quad\text{ and }\quad b_0 > b_1  > \cdots > b_n > 0$$ If $N$ is the number of sign changes in the coefficients $a_j$, then the number of real roots of $P(x)$ is bounded by $N$.

It is easy to see the sum you have has exactly $n$ sign changes. This means it has at most $n$ solutions. Since we have located $n$ solutions
for the problem, that's all the solutions it has.
Notes/Refs

*

*$\color{blue}{[1]}$ - For a modern introduction to this subject, I'll recommned
G.J.O Jameson, Counting zeros of generalized polynomials: Descartes' rule of signs and Laguerre's extensions, (Math. Gazette 90, no. 518 (2006), 223-234).
An online copy can be found here.
