There are two ways we can go about this. We can either take a closer look at the Lindemann theorem, or we can talk about the definition of “transcendental number”. I’m not going to put a full proof here.
If you look at the Lindemann theorem, you can transform the equation so that it uses π instead of e. Multiply all of the exponents by i (which is algebraic!) and then use Euler’s identity. You end up with the same formula, but with π instead of e.
However, if we already know that π is transcendental (which is proven by the Lindemann theorem using the above technique), we can rewrite any linear combination of B = {1, π, π², π³, …} as P(π) where P is a polynomial with coefficients in ℚ. Because π is transcendental, we know that P(π)=0 only if P is the zero polynomial (that is the definition of transcendental number).
In general, one of the big tricks here is that the set of polynomials is a vector space, and the powers B = {1, x, x², x³, …} span the entire vector space.
Do we know that? My search doesn’t get more than https://www.encyclopediaofmath.org/index.php/Lindemann_theor..., which proves it for “𝑒, 𝑒², 𝑒³, …“.