The recursion defines a unique monic ${\mathrm{\Phi }}_n\in \mathbb{Z}[t]$, of degree $\phi (n)$

The assertion to be proved by strong induction on $n$ is: the division defining ${\mathrm{\Phi }}_n$ is exact, ${\mathrm{\Phi }}_n\in \mathbb{Z}[t]$ is monic, ${\prod }_{d\mid n}{\mathrm{\Phi }}_d=t^n-1$, and $\mathrm{deg}{\mathrm{\Phi }}_n=\phi (n)$. At $n=1$ the recursion sets ${\mathrm{\Phi }}_1=t-1$ outright, the only positive divisor of $1$ is $1$ so the product is ${\mathrm{\Phi }}_1=t-1$, and $\mathrm{deg}{\mathrm{\Phi }}_1=1=\phi (1)$.

Inductive hypothesis: fix $n\ge 2$ and assume the assertion for every $m$ with $1\le m<n$.

Since $\mathrm{char}\mathbb{Q}=0$ does not divide $n$, [L2] and [L3] supply a splitting field $E$ of $t^n-1$ over $\mathbb{Q}$ in which $t^n-1$ is separable and ${\mu }_n(E)$ is cyclic of order $n$. For a positive divisor $d$ of $n$ put $S_d:=\{\zeta \in {\mu }_n(E):\mathrm{ord}(\zeta )=d\}$ and ${\mathrm{\Psi }}_d:={\prod }_{\zeta \in S_d}(t-\zeta )\in E[t]$, a monic polynomial of degree $\mathrm{\mid }{S}_d\mathrm{\mid }$.

For every positive divisor $m$ of $n$ one has $t^m-1={\prod }_{\zeta \in {\mu }_m(E)}(t-\zeta )$ with $\mid {\mu }_m(E)\mid =m$: the polynomial $t^m-1$ divides $t^n-1$ in $\mathbb{Z}[t]$, since $t^n-1=(t^m-1)(t^{n-m}+t^{n-2m}+\cdots +1)$ when $m\mid n$, so it splits over $E$ by [L3], and a repeated root of $t^m-1$ in an extension would be a repeated root of $t^n-1$ there, which [L4] excludes; hence its $m$ roots are distinct and they are by definition the elements of ${\mu }_m(E)$.

For every positive divisor $m$ of $n$, ${\mu }_m(E)$ is the disjoint union of the $S_d$ over positive divisors $d$ of $m$: an element $\zeta \in {\mu }_n(E)$ has finite order dividing $n$, and ${\zeta }^m=1$ holds exactly when $\mathrm{ord}(\zeta )\mid m$ by [L5]. Hence ${\prod }_{d\mid m}{\mathrm{\Psi }}_d={\prod }_{\zeta \in {\mu }_m(E)}(t-\zeta )=t^m-1$ by step 2.1.

For every positive divisor $d$ of $n$ with $d<n$ one has ${\mathrm{\Phi }}_d={\mathrm{\Psi }}_d$, by induction on $d$ through the divisors of $n$: at $d=1$ both equal $t-1$, since $S_1=\{1\}$; and if ${\mathrm{\Phi }}_e={\mathrm{\Psi }}_e$ for every positive divisor $e$ of $d$ with $e<d$, then step 1.2 and step 3.1 give $\left({\prod }_{e\mid d,e<d}{\mathrm{\Psi }}_e\right){\mathrm{\Phi }}_d=t^d-1=\left({\prod }_{e\mid d,e<d}{\mathrm{\Psi }}_e\right){\mathrm{\Psi }}_d$, and cancelling the nonzero left factor in the integral domain $E[t]$ ([L7]) yields ${\mathrm{\Phi }}_d={\mathrm{\Psi }}_d$.

Write $P:={\prod }_{d\mid n,d<n}{\mathrm{\Phi }}_d$, monic in $\mathbb{Z}[t]$ by step 1.2 and [L7]. By step 4.1 and step 3.1 applied with $m=n$, in $E[t]$ one has $t^n-1=\left({\prod }_{d\mid n,d<n}{\mathrm{\Psi }}_d\right){\mathrm{\Psi }}_n=P{\mathrm{\Psi }}_n$.

The division of $t^n-1$ by $P$ in $\mathbb{Z}[t]$ is exact and its quotient is ${\mathrm{\Psi }}_n$: by [L1] over $\mathbb{Z}$ there are unique $q,r\in \mathbb{Z}[t]$ with $t^n-1=qP+r$ and $r=0$ or $\mathrm{deg}r<\mathrm{deg}P$; this is also a division by the monic $P$ in $E[t]$, where step 5.1 exhibits the division with quotient ${\mathrm{\Psi }}_n$ and remainder $0$, so the uniqueness clause of [L1] over $E$ forces $q={\mathrm{\Psi }}_n$ and $r=0$. Hence ${\mathrm{\Phi }}_n=q={\mathrm{\Psi }}_n$ is monic in $\mathbb{Z}[t]$ and ${\prod }_{d\mid n}{\mathrm{\Phi }}_d=P{\mathrm{\Phi }}_n=t^n-1$.

Degrees: taking degrees in $t^n-1=P{\mathrm{\Phi }}_n$ with [L7] gives $n=\mathrm{deg}P+\mathrm{deg}{\mathrm{\Phi }}_n$, and $\mathrm{deg}P={\sum }_{d\mid n,d<n}\mathrm{deg}{\mathrm{\Phi }}_d={\sum }_{d\mid n,d<n}\phi (d)$ by step 1.2 and [L7]; so $\mathrm{deg}{\mathrm{\Phi }}_n=n-{\sum }_{d\mid n,d<n}\phi (d)=\phi (n)$ by [L6].

This is the assertion at $n$, so the strong induction is complete and the assertion holds for every $n\ge 1$. Uniqueness of the family follows by the same induction: if $({\mathrm{\Phi }}_m^{\mathrm{\prime }}{)}_{m\ge 1}$ is monic in $\mathbb{Z}[t]$ with ${\prod }_{d\mid m}{\mathrm{\Phi }}_d^{\mathrm{\prime }}=t^m-1$ for all $m$, then ${\mathrm{\Phi }}_1^{\mathrm{\prime }}=t-1={\mathrm{\Phi }}_1$, and if ${\mathrm{\Phi }}_m^{\mathrm{\prime }}={\mathrm{\Phi }}_m$ for all $m<n$ then $P{\mathrm{\Phi }}_n^{\mathrm{\prime }}=t^n-1=P{\mathrm{\Phi }}_n$ with $P\mathrm{\ne }0$ in the integral domain $\mathbb{Z}[t]$ ([L7]), so ${\mathrm{\Phi }}_n^{\mathrm{\prime }}={\mathrm{\Phi }}_n$.