fix: use correct ES6 module syntax

- Turns out `export default` is not the same as `modules.export = xyz`
- The actual equivilant is `export = xyz` with `esModuleInterop` enabled
- This is a better form than `modules.export = xyz` because we are able
to use the type system
- See: https://stackoverflow.com/questions/40294870/module-exports-vs-export-default-in-node-js-and-es6 and https://www.typescriptlang.org/docs/handbook/migrating-from-javascript.html#weeding-out-errors

Author: Sean Dawson

src/index.ts

@@ -1,15 +1,15 @@
 import * as path from 'path'
 import * as fs from 'fs-extra'
 import * as _ from 'lodash'
-import * as globby from 'globby'
+import globby from 'globby'
 
 import * as typescript from './typescript'
 import { watchFiles } from './watchFiles'
 
 const SERVERLESS_FOLDER = '.serverless'
 const BUILD_FOLDER = '.build'
 
-export class TypeScriptPlugin {
+class TypeScriptPlugin {
   private originalServicePath: string
   private isWatching: boolean
 
@@ -278,4 +278,4 @@ export class TypeScriptPlugin {
   }
 }
 
-export default TypeScriptPlugin
+export = TypeScriptPlugin

tsconfig.json

@@ -4,7 +4,8 @@
     "rootDir": ".",
     "target": "es6",
     "sourceMap": true,
-    "outDir": "dist"
+    "outDir": "dist",
+    "esModuleInterop": true
   },
   "exclude": [
     "node_modules"