Type-Level Exponentiation in TypeScript

This challenge involves implementing exponentiation directly at the TypeScript type level. You'll be creating a type that can calculate Base ^ Exponent for non-negative integer Base and Exponent types. This is a fundamental exercise in exploring the expressive power of TypeScript's advanced type manipulation, useful for building complex type-safe libraries and domain-specific languages.

Problem Description

Your task is to define a TypeScript type TypeLevelExponentiation<Base extends number, Exponent extends number> that computes the result of Base raised to the power of Exponent at compile time.

Key Requirements:

• The Base and Exponent will be positive integers (represented as number literals in TypeScript types).
• The type should handle the base case where Exponent is 0, in which case the result is 1.
• The type should recursively apply multiplication for positive exponents.
• Consider edge cases like Base = 0 and Exponent = 0.

Expected Behavior:

The type TypeLevelExponentiation<B, E> should resolve to a number literal representing B ^ E.

Edge Cases to Consider:

• TypeLevelExponentiation<any, any>: While the problem focuses on number literals, think about how the type might behave with any. (For this challenge, assume valid number literals).
• Large exponents: TypeScript's recursive depth can be a limitation. For this challenge, focus on correctness for reasonably sized exponents that won't hit compiler limits.
• Negative exponents: Not required for this challenge. Assume Exponent >= 0.
• Non-integer bases/exponents: Not required. Assume integer literals.

Examples

Example 1:

type Result1 = TypeLevelExponentiation<2, 3>;
// Expected: Result1 should be type 8

Explanation: 2 ^ 3 is calculated as 2 * 2 * 2 = 8.

Example 2:

type Result2 = TypeLevelExponentiation<5, 0>;
// Expected: Result2 should be type 1

Explanation: Any non-zero number raised to the power of 0 is 1.

Example 3:

type Result3 = TypeLevelExponentiation<0, 5>;
// Expected: Result3 should be type 0

Explanation: 0 raised to any positive power is 0.

Example 4:

type Result4 = TypeLevelExponentiation<1, 10>;
// Expected: Result4 should be type 1

Explanation: 1 raised to any power is 1.

Example 5:

type Result5 = TypeLevelExponentiation<0, 0>;
// Expected: Result5 should be type 1

Explanation: Conventionally, 0^0 is considered 1 in many mathematical contexts, particularly in combinatorics and polynomial expansions.

Constraints

• Base and Exponent will be non-negative integer number literals (e.g., 0, 1, 5, 10).
• The solution should only use TypeScript's type system features. No runtime code is involved.
• Focus on correctness for exponents up to a reasonable limit (e.g., 10-15) to avoid hitting TypeScript's recursion depth limits.

Notes

This problem requires understanding and applying recursive type definitions. You'll likely need a helper type to manage the multiplication process. Consider how to decrement the exponent in each recursive step and how to accumulate the result. The base case of Exponent = 0 is crucial for terminating the recursion.