I was interested in the proof for this fact because it is used to prove M$\ddot o$bius Inversion Formula. However, I did not completely understand how proof wiki used a series of binomial coefficients to prove the fact. So I tried to find another proof.
$\mathrm Objective$: Prove $$(\mu*1)=\delta_1$$ using $$\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}$$ and $$\frac{1}{\zeta(s)}=\sum_{k=1}^\infty \frac{\mu(k)}{k^s}$$
$\mathrm Work$: $$\zeta(s)\frac{1}{\zeta(s)}= \sum_{n=1}^\infty \frac{1}{n^s}\sum_{k=1}^\infty \frac{\mu(k)}{k^s}$$ $$1=\sum_{n=1}^\infty\sum_{k=1}^\infty \frac{\mu(k)}{n^sk^s}$$ Let $m=nk$ $$1=\sum_{m=1}^\infty\sum_{nk=m} \frac{\mu(k)}{m^s}$$ $$1=\sum_{m=1}^\infty \frac{\mu*1}{n^s}$$ So $$(\mu*1)= 1, n=1$$ And $$(\mu*1)= 0, n\neq1$$ Which is the definition of Kronecker Delta $\delta_1$.
$\mathrm Question$:Was my work valid? I am still in high school so I doubt that my calculus teacher can help me.
Never doubt your Calculus teacher.
Jokes apart, your work is fine, but let us recall a few facts for future reference, too.
Definition. A function $f:\mathbb{N}^+\to\mathbb{C}$ is said to be multiplicative if $\gcd(n,m)=1$ ensures $f(n)\,f(m)=f(nm)$. Completely (or totally) multiplicative if $f(n)\,f(m)=f(mn)$ holds in any case.
Lemma. The values of a multiplicative function are fixed by the values that such function attains over prime powers, by the fundamental theorem of Arithmetics ($\mathbb{Z}$ is a $\text{UFD}$).
Definition. The Dirichlet convolution/multiplicative convolution between two arithmetic functions is defined through $(f*g)(n)=\sum_{d \mid n}f(d)\,g\left(\tfrac{n}{d}\right)$. The operator $*$ is both associative and commutative, and the convolution between two multiplicative functions still is a multiplicative function.
Definition. Any multiplicative function $\chi$ is associated with a Dirichlet L-function $$ L(\chi,s) = \sum_{n\geq 1}\frac{\chi(n)}{n^s} $$ and assuming that $\chi$ takes values in $\{z\in\mathbb{C}:|z|\leq 1\}$ the last series is absolutely convergent over the region $\text{Re}\, s>1$. The region of conditional convergence might be larger, depending on $\chi$.
Lemma. We have the formal identity $$ L(f*g,s) = L(f,s)\cdot L(g,s) $$ which is an actual identity over the region $\text{Re}\,s>1$.
Theorem (Euler's product). For any $s$ such that $\text{Re}\,s>1$ we have: $$ L(\chi,s) = \prod_{p}\left(1+\frac{\chi(p)}{p^s}+\frac{\chi(p^2)}{p^{2s}}+\frac{\chi(p^3)}{p^{3s}}+\ldots\right)$$ which simplifies into $$ L(\chi,s) = \prod_{p}\left(1-\frac{\chi(p)}{p^s}\right)^{-1} $$ if $\chi$ is completely multiplicative.
Theorem (Principle of analytic continuation). If $L(\chi_1,s)=L(\chi_2,s)$ holds over some region of the form $\text{Re}\,s>c$, then $\chi_1\equiv \chi_2$. This can be proved by applying $\mathcal{L}^{-1}$ (the inverse Laplace transform) to both sides, then by invoking the principle of analytic continuation in its standard form.
Given this wonderful intro, we are ready to give two proofs of the Moebius inversion formula.
We recall that $\mu(n)$ differs from zero iff $n$ is a square-free integer, and in such a case it equals $(-1)^{\omega(n)}$, i.e. $\pm 1$ according to the parity of the number of prime divisors of $n$. $\mu(n)$ is obviously a multiplicative function, like the function which constantly equals one. By the machinery above, in order to prove $\mu * 1 = \delta_1$ it is enough to prove $L(\mu,s)\cdot L(1,s) = L(\delta_1,s)=1 $ for any $s$ with a large enough real part. On the other hand such identity is trivial by Euler's product, since
$$ L(\mu,s)=\prod_{p}\left(1-\frac{1}{p^s}\right),\qquad L(1,s)=\zeta(s)=\prod_{p}\left(1-\frac{1}{p^s}\right)^{-1}.$$ This works like a charm, but it is definitely an overkill. Since $\mu*1$ is a multiplicative function, in order to prove $\mu*1=\delta_1$ it is enough to show that $(\mu*1)(p^k)=0$ holds for any prime $p$ and any exponent $k\geq 1$. By definition of $*$ we have
$$ (\mu*1)(p^k) = \sum_{d\mid p^k} \mu(d) = \sum_{h=0}^{1}\mu(p^h) = 1-1 = 0 $$ and we are equally done.