Proof:

To prove that $6^n - 5n$ always leaves a remainder of 1 when divided by 25, we will use the binomial theorem.

Binomial Theorem:
The binomial theorem states that for any positive integer n, the expansion of $(a + b)^n$ can be expressed as:

$(a + b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1} b^1 + \binom{n}{2}a^{n-2} b^2 + ... + \binom{n}{n-1}a^1 b^{n-1} + \binom{n}{n}a^0 b^n$

where $\binom{n}{k}$ represents the binomial coefficient, which is defined as:

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$

Using the Binomial Theorem:

Let's start by expanding $(6+(-5))^n$ using the binomial theorem. Since $(-5n)$ is equal to $(-5)\cdot(1)n$, we can write:

$(6-5n)^n = \binom{n}{0}6^n (-5n)^0 + \binom{n}{1}6^{n-1} (-5n)^1 + \binom{n}{2}6^{n-2} (-5n)^2 + ... + \binom{n}{n-1}6^1 (-5n)^{n-1} + \binom{n}{n}6^0 (-5n)^n$

Simplifying each term, we have:

$\binom{n}{0}6^n (-5n)^0 = 6^n$

$\binom{n}{1}6^{n-1} (-5n)^1 = -5n\cdot6^{n-1}$

$\binom{n}{2}6^{n-2} (-5n)^2 = 10n^2\cdot6^{n-2}$

Continuing this pattern, the general term can be written as:

$\binom{n}{k}6^{n-k} (-5n)^k = (-1)^k\cdot\binom{n}{k} \cdot6^{n-k} \cdot n^k \cdot 5^k$

Proving the Remainder:

Now, let's consider the terms in the expansion, except for the first term $6^n$:

$S = -5n\cdot6^{n-1} + 10n^2\cdot6^{n-2} - 10n^3\cdot6^{n-3} + ... + (-1)^k\cdot\binom{n}{k} \cdot6^{n-k} \cdot n^k \cdot 5^k$

We can observe that every term in this sum is divisible by 25, except for the first term $-5n\cdot6^{n-1}$. Therefore, the remainder when $6^n - 5n$ is divided by 25 is equal to the remainder when $-5n\cdot6^{n-1}$ is divided by 25.

Proving the Remainder is