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One can make use of Simplify with Assumptions

I. Compute the sum

s=Sum[HarmonicNumber[n,5]/n^8,{n,1,Infinity}]
(* -(1/63) π^6 Zeta[7]-13/15 π^4 Zeta[9]-55 π^2 Zeta[11]+644 Zeta[13] *)

II. Make a table of Zeta-functions with even arguments

t=Flatten[Table[{ζ[2n]==Zeta[2n]},{n,0,6}]]
(* {ζ[0]==-(1/2),ζ[2]==π^2/6,ζ[4]==π^4/90,ζ[6]==π^6/945,ζ[8]==π^8/9450,ζ[10]==π^10/93555,ζ[12]==(691 π^12)/638512875} *)

III. Simplify with assumptions

Simplify[s/.{π->x,Zeta[n_]->ζ[n]},Assumptions->t/.{π->x}]
(* -15 ζ[6] ζ[7]-78 ζ[4] ζ[9]-330 ζ[2] ζ[11]+644 ζ[13] *)

Comments:

1. A much cleaner solution would be to use the method of Carl Woll to deactivate the Zetafunction. However, it only works if the sum can be computed in the desired terms, i.e., in terms of Zeta[2n+1] and Zeta[2n]. This seems not to be the case.
2. The present method only replaces the occurrences of $\pi^{2n}$ for $n\ge1$. The zeroth-order term is left unchanged. Would be nice if someone can come up with a better solution that also takes Zeta[0] into account.

One can make use of Simplify with Assumptions

I. Compute the sum

s=Sum[HarmonicNumber[n,5]/n^8,{n,1,Infinity}]
(* -(1/63) π^6 Zeta[7]-13/15 π^4 Zeta[9]-55 π^2 Zeta[11]+644 Zeta[13] *)

II. Make a table of Zeta-functions with even arguments

t=Flatten[Table[{ζ[2n]==Zeta[2n]},{n,0,6}]]
(* {ζ[0]==-(1/2),ζ[2]==π^2/6,ζ[4]==π^4/90,ζ[6]==π^6/945,ζ[8]==π^8/9450,ζ[10]==π^10/93555,ζ[12]==(691 π^12)/638512875} *)

III. Simplify with assumptions

Simplify[s/.{π->x,Zeta[n_]->ζ[n]},Assumptions->t/.{π->x}]
(* -15 ζ[6] ζ[7]-78 ζ[4] ζ[9]-330 ζ[2] ζ[11]+644 ζ[13] *)

Comments:

1. A much cleaner solution would be to use the method of Carl Woll to deactivate the Zetafunction. However, it only works if the sum can be computed in the desired terms, i.e., in terms of Zeta[2n+1] and Zeta[2n]. This seems not to be the case.
2. The present method only replaces the occurrences of $\pi^{2n}$ for $n\ge1$. The zeroth-order term is left unchanged. Below I show how Zeta[0] can be taken into account.

Different approach using SolveAlways

Define some functions

Clear[sf,sx,sxx]
sf[a_,b_]:=Sum[HarmonicNumber[n,a]/n^b,{n,1,Infinity}]/.{Zeta[n_]->ζ[n]}
sxx[c_]:=Sum[a[i]ζ[i]Zeta[c-i],{i,c,Floor[c/2],-2}]
sx[c_]:=Sum[a[i]ζ[i]ζ[c-i],{i,c,Floor[c/2],-2}]

Let us try

k=5;l=8;
sx[k+l]/.First[SolveAlways[(sf[k,l]==sxx[k+l]),Table[ζ[i],{i,k+l,Floor[(k+l)/2],-2}]]]
(* -15 ζ[6] ζ[7]-78 ζ[4] ζ[9]-330 ζ[2] ζ[11]-1288 ζ[0] ζ[13] *)

One can make use of Simplify with Assumptions

I. Compute the sum

s=Sum[HarmonicNumber[n,5]/n^8,{n,1,Infinity}]
(* -(1/63) π^6 Zeta[7]-13/15 π^4 Zeta[9]-55 π^2 Zeta[11]+644 Zeta[13] *)

II. Make a table of Zeta-functions with even arguments

t=Flatten[Table[{ζ[2n]==Zeta[2n]},{n,0,6}]]
(* {ζ[0]==-(1/2),ζ[2]==π^2/6,ζ[4]==π^4/90,ζ[6]==π^6/945,ζ[8]==π^8/9450,ζ[10]==π^10/93555,ζ[12]==(691 π^12)/638512875} *)

III. Simplify with assumptions

Simplify[s/.{π->x,Zeta[n_]->ζ[n]},Assumptions->t/.{π->x}]
(* -15 ζ[6] ζ[7]-78 ζ[4] ζ[9]-330 ζ[2] ζ[11]+644 ζ[13] *)

Comments:

1. A much cleaner solution would be to use the method of Carl Woll to deactivate the Zetafunction. However, it only works if the sum can be computed in the desired terms, i.e., in terms of Zeta[2n+1] and Zeta[2n]. This seems not to be the case.
2. The present method only replaces the occurrences of $\pi^{2n}$ for $n\ge1$. Zeroth-order term is left unchanged.

One can make use of Simplify with Assumptions

I. Compute the sum

s=Sum[HarmonicNumber[n,5]/n^8,{n,1,Infinity}]
(* -(1/63) π^6 Zeta[7]-13/15 π^4 Zeta[9]-55 π^2 Zeta[11]+644 Zeta[13] *)

II. Make a table of Zeta-functions with even arguments

t=Flatten[Table[{ζ[2n]==Zeta[2n]},{n,0,6}]]
(* {ζ[0]==-(1/2),ζ[2]==π^2/6,ζ[4]==π^4/90,ζ[6]==π^6/945,ζ[8]==π^8/9450,ζ[10]==π^10/93555,ζ[12]==(691 π^12)/638512875} *)

III. Simplify with assumptions

Simplify[s/.{π->x,Zeta[n_]->ζ[n]},Assumptions->t/.{π->x}]
(* -15 ζ[6] ζ[7]-78 ζ[4] ζ[9]-330 ζ[2] ζ[11]+644 ζ[13] *)

Comments:

1. A much cleaner solution would be to use the method of Carl Woll to deactivate the Zetafunction. However, it only works if the sum can be computed in the desired terms, i.e., in terms of Zeta[2n+1] and Zeta[2n]. This seems not to be the case.
2. The present method only replaces the occurrences of $\pi^{2n}$ for $n\ge1$. The zeroth-order term is left unchanged. Would be nice if someone can come up with a better solution that also takes Zeta[0] into account.

One can make use of Simplify with Assumptions

I. Compute the sum

s=Sum[HarmonicNumber[n,5]/n^8,{n,1,Infinity}]
(* -(1/63) π^6 Zeta[7]-13/15 π^4 Zeta[9]-55 π^2 Zeta[11]+644 Zeta[13] *)

II. Table of Zeta-function of even arguments

t=Flatten[Table[{ζ[2n]==Zeta[2n]},{n,0,6}]]
(* {ζ[0]==-(1/2),ζ[2]==π^2/6,ζ[4]==π^4/90,ζ[6]==π^6/945,ζ[8]==π^8/9450,ζ[10]==π^10/93555,ζ[12]==(691 π^12)/638512875} *)

III. Simplify with assumptions

Simplify[s/.{π->x,Zeta[n_]->ζ[n]},Assumptions->t/.{π->x}]
(* -15 ζ[6] ζ[7]-78 ζ[4] ζ[9]-330 ζ[2] ζ[11]+644 ζ[13] *)

One can make use of Simplify with Assumptions

I. Compute the sum

s=Sum[HarmonicNumber[n,5]/n^8,{n,1,Infinity}]
(* -(1/63) π^6 Zeta[7]-13/15 π^4 Zeta[9]-55 π^2 Zeta[11]+644 Zeta[13] *)

II. Make a table of Zeta-functions with even arguments

t=Flatten[Table[{ζ[2n]==Zeta[2n]},{n,0,6}]]
(* {ζ[0]==-(1/2),ζ[2]==π^2/6,ζ[4]==π^4/90,ζ[6]==π^6/945,ζ[8]==π^8/9450,ζ[10]==π^10/93555,ζ[12]==(691 π^12)/638512875} *)

III. Simplify with assumptions

Simplify[s/.{π->x,Zeta[n_]->ζ[n]},Assumptions->t/.{π->x}]
(* -15 ζ[6] ζ[7]-78 ζ[4] ζ[9]-330 ζ[2] ζ[11]+644 ζ[13] *)

Comments:

1. A much cleaner solution would be to use the method of Carl Woll to deactivate the Zetafunction. However, it only works if the sum can be computed in the desired terms, i.e., in terms of Zeta[2n+1] and Zeta[2n]. This seems not to be the case.
2. The present method only replaces the occurrences of $\pi^{2n}$ for $n\ge1$. Zeroth-order term is left unchanged.